Aula 2 Radioatividade

Prévia do material em texto

RADIOATIVIDADE E 
FONTES RADIOATIVAS
FONTES RADIOATIVAS
! “Transformação espontânea, em um átomo instável, que 
resulta em um novo elemento.” 
! Essa transformação: 
! Independe do estado físico 
! Gás, líquido ou sólido 
! Independe do estado químico 
! Independe de pressão, temperatura, umidade...
NOMECLATURA
Isótopos 
Mesmo número atômico e diferente número de 
nêutrons
H
1
1 H
1
2 H
1
3
NOMECLATURA
Isótopos 
Mesmo número atômico e diferente número de 
nêutrons
H
1
1 H
1
2 H
1
3
N
ZH-1
H-2
H-3
He-3
He-4
He-5
Li-4
Li-5
Li-6
Li-7
...
n
0 1 2 3
1
0
2
3
4
5
4
# nêutrons
INSTABILIDADE DO NÚCLEO
! Isótopos Estáveis (apenas os de cor preto na tabela) 
! Z par e N par = 162 isótopos estáveis 
! Z par e N ímpar ou 
! Z ímpar e N par = 108 isótopos estáveis 
! Z ímpar e N ímpar = 4 isótopos estáveis
56 3 The Nucleus and Nuclear Radiation
different isotopes of the same element are responsible for hyperfine structure in
the spectra of elements. As we mentioned at the end of Section 2.8, the existence
of isotopes has a big effect on the vibration–rotation spectra of molecules.
Nucleons are bound together in a nucleus by the action of the strong, or nuclear,
force. The range of this force is only of the order of nuclear dimensions, ∼10–15 m,
and it is powerful enough to overcome the Coulomb repulsion of the protons in
the nucleus. Figure 3.1(a) schematically shows the potential energy of a proton as a
function of the distance r separating its center and the center of a nucleus. The po-
tential energy is zero at large separations. As the proton comes closer, its potential
energy increases, due to the work done against the repulsive Coulomb force that
Fig. 3.1 (a) Potential energy (PE) of a proton as a function of
its separation r from the center of a nucleus, (b) Potential
energy of a neutron and a nucleus as a function of r. The
uncharged neutron has no repulsive Coulomb barrier to
overcome when approaching a nucleus.
Força Forte ou Força Nuclear Forte: 
 mantém os prótons e nêutrons juntos
Repulsão das cargas positivas
Para prótons
56 3 The Nucleus and Nuclear Radiation
different isotopes of the same element are responsible for hyperfine structure in
the spectra of elements. As we mentioned at the end of Section 2.8, the existence
of isotopes has a big effect on the vibration–rotation spectra of molecules.
Nucleons are bound together in a nucleus by the action of the strong, or nuclear,
force. The range of this force is only of the order of nuclear dimensions, ∼10–15 m,
and it is powerful enough to overcome the Coulomb repulsion of the protons in
the nucleus. Figure 3.1(a) schematically shows the potential energy of a proton as a
function of the distance r separating its center and the center of a nucleus. The po-
tential energy is zero at large separations. As the proton comes closer, its potential
energy increases, due to the work done against the repulsive Coulomb force that
Fig. 3.1 (a) Potential energy (PE) of a proton as a function of
its separation r from the center of a nucleus, (b) Potential
energy of a neutron and a nucleus as a function of r. The
uncharged neutron has no repulsive Coulomb barrier to
overcome when approaching a nucleus.
56 3 The Nucleus and Nuclear Radiation
different isotopes of the same element are responsible for hyperfine structure in
the spectra of elements. As we mentioned at the end of Section 2.8, the existence
of isotopes has a big effect on the vibration–rotation spectra of molecules.
Nucleons are bound together in a nucleus by the action of the strong, or nuclear,
force. The range of this force is only of the order of nuclear dimensions, ∼10–15 m,
and it is powerful enough to overcome the Coulomb repulsion of the protons in
the nucleus. Figure 3.1(a) schematically shows the potential energy of a proton as a
function of the distance r separating its center and the center of a nucleus. The po-
tential energy is zero at large separations. As the proton comes closer, its potential
energy increases, due to the work done against the repulsive Coulomb force that
Fig. 3.1 (a) Potential energy (PE) of a proton as a function of
its separation r from the center of a nucleus, (b) Potential
energy of a neutron and a nucleus as a function of r. The
uncharged neutron has no repulsive Coulomb barrier to
overcome when approaching a nucleus.
Força Forte ou Força Nuclear Forte: 
 mantém os prótons e nêutrons juntos
Repulsão das cargas positivas
Para nêutrons
Para prótons
Níveis de Energia dos 
nucleons (dentro do núcleo)
2, 8, 20, 28, 50, 82, and 126
http://oeis.org/A018226
The stability of those nuclei with magic numbers of neutrons makes them less likely 
to be excited by neutron bombardment. The probability for absorption of an 
incident is expressed as an effective cross-section which is presented by the target 
nucleus to those incoming neutrons. The common unit for cross-sections is the barn, 
and the vertical axis on the illustration is labeled in millibarns. 
Seção de choque para
excitação
NÚMEROS MÁGICOS
! Núcleos com 2 , 8, 20, 50, 82 e 126 nucleons 
(N ou Z) são mais estáveis 
! Princípio análogo ao modelo atômico 
! Goeppert-Mayer e Hans Jensen (nobel de 
1963) 
ENERGIA DE LIGAÇÃO
ABUNDÂNCIA 
RELATIVA
Nuclei which have neutron number and proton (atomic) numbers each equal to one of the 
magic numbers are called "double magic", and are especially stable against decay. Examples 
of double magic isotopes include helium-4, oxygen-16, calcium-40, nickel-48, nickel-78, 
and lead-208.
Double-magic effects may allow existence of stable isotopes which otherwise would not have 
been expected. An example is calcium-40, with 20 neutrons and 20 protons, which is the 
heaviest stable isotope made of the same number of protons and neutrons. 
Both calcium-48 and nickel-48 are double magic because calcium-48 has 20 protons and 
28 neutrons while nickel-48 has 28 protons and 20 neutrons. Calcium-48 is very neutron-rich 
for such a light element, but like calcium-40, it is made stable by being double magic. 
Nickel-48, discovered in 1999, is the most proton-rich isotope known beyond helium-3.[5] 
At the other extreme, nickel-78 is also doubly magical, with 28 protons and 50 neutrons, 
a ratio observed only in much heavier elements (Ni-78: 28/50 = 0.56; U-238: 92/146 = 0.63).
[6]
EXEMPLOS
N=17Z=15
... ...
P32
15
N=16Z=16
S32
16
ENERGIAS ASSOCIADAS ÀS 
TRANSFORMAÇÕES NUCLEARES
As energias associadas à transformações nucleares são da ordem de 
106 eV (ou MeV). 
3.2 Nuclear Binding Energies 59
greater than the energies associated with the valence electrons that are involved in
chemical reactions. This factor characterizes the enormous difference in the en-
ergy released when an atom undergoes a nuclear transformation as compared with
a chemical reaction.
The energy associated with exothermic nuclear reactions comes from the con-
version of mass into energy. If the mass loss is !M, then the energy released, Q ,
is given by Einstein’s relation, Q= (!M)c2, where c is the velocity of light. In this
section we discuss the energetics of nuclear transformations.
We first establish the quantitative relationship between atomic mass units
(AMU) and energy (MeV). By definition, the 12C atom has a mass of exactly
12 AMU. Since its gram atomic weight is 12 g, it follows that
1 AMU= 1/(6.02× 1023)= 1.66× 10–24 g= 1.66× 10–27 kg. (3.4)
Using the Einstein relation and c= 3× 108 m s–1, we obtain
1 AMU = (1.66× 10–27)(3× 108)2
= 1.49× 10–10 J (3.5)
= 1.49× 10
–10 J
1.6× 10–13 J MeV–1 = 931 MeV. (3.6)
More precisely, 1 AMU = 931.49 MeV.
We now consider one of the simplestnuclear reactions, the absorption of a ther-
mal neutron by a hydrogen atom, accompanied by emission of a gamma ray. This
reaction, which is very important for understanding the thermal-neutron dose to
the body, can be represented by writing
1
0n +
1
1H→ 21H + 00γ, (3.7)
the photon having zero charge and mass. The reaction can also be designated
1
1H(n,γ)
2
1H. To find the energy released, we compare the total masses on both sides
of the arrow. Appendix D contains data on nuclides which we shall frequently use.
The atomic weight M of a nuclide of mass number A can be found from the mass
difference, !, given in column 3. The quantity !=M – A gives the difference be-
tween the nuclide’s atomic weight and its atomic mass number, expressed in MeV.
(By definition, != 0 for the 12C atom.) Since we are interested only in energy dif-
ferences in the reaction (3.7), we obtain the energy released, Q, directly from the
values of !, without having to calculate the actual masses of the neutron and indi-
vidual atoms. Adding the ! values for 10n and
1
1H and subtracting that for
2
1H, we
find
Q= 8.0714 + 7.2890 – 13.1359= 2.2245 MeV. (3.8)
This energy appears as a gamma photon emitted when the capture takes place (the
thermal neutron has negligible kinetic energy).
3.2 Nuclear Binding Energies 59
greater than the energies associated with the valence electrons that are involved in
chemical reactions. This factor characterizes the enormous difference in the en-
ergy released when an atom undergoes a nuclear transformation as compared with
a chemical reaction.
The energy associated with exothermic nuclear reactions comes from the con-
version of mass into energy. If the mass loss is !M, then the energy released, Q ,
is given by Einstein’s relation, Q= (!M)c2, where c is the velocity of light. In this
section we discuss the energetics of nuclear transformations.
We first establish the quantitative relationship between atomic mass units
(AMU) and energy (MeV). By definition, the 12C atom has a mass of exactly
12 AMU. Since its gram atomic weight is 12 g, it follows that
1 AMU= 1/(6.02× 1023)= 1.66× 10–24 g= 1.66× 10–27 kg. (3.4)
Using the Einstein relation and c= 3× 108 m s–1, we obtain
1 AMU = (1.66× 10–27)(3× 108)2
= 1.49× 10–10 J (3.5)
= 1.49× 10
–10 J
1.6× 10–13 J MeV–1 = 931 MeV. (3.6)
More precisely, 1 AMU = 931.49 MeV.
We now consider one of the simplest nuclear reactions, the absorption of a ther-
mal neutron by a hydrogen atom, accompanied by emission of a gamma ray. This
reaction, which is very important for understanding the thermal-neutron dose to
the body, can be represented by writing
1
0n +
1
1H→ 21H + 00γ, (3.7)
the photon having zero charge and mass. The reaction can also be designated
1
1H(n,γ)
2
1H. To find the energy released, we compare the total masses on both sides
of the arrow. Appendix D contains data on nuclides which we shall frequently use.
The atomic weight M of a nuclide of mass number A can be found from the mass
difference, !, given in column 3. The quantity !=M – A gives the difference be-
tween the nuclide’s atomic weight and its atomic mass number, expressed in MeV.
(By definition, != 0 for the 12C atom.) Since we are interested only in energy dif-
ferences in the reaction (3.7), we obtain the energy released, Q, directly from the
values of !, without having to calculate the actual masses of the neutron and indi-
vidual atoms. Adding the ! values for 10n and
1
1H and subtracting that for
2
1H, we
find
Q= 8.0714 + 7.2890 – 13.1359= 2.2245 MeV. (3.8)
This energy appears as a gamma photon emitted when the capture takes place (the
thermal neutron has negligible kinetic energy).
3.2 Nuclear Binding Energies 59
greater than the energies associated with the valence electrons that are involved in
chemical reactions. This factor characterizes the enormous difference in the en-
ergy released when an atom undergoes a nuclear transformation as compared with
a chemical reaction.
The energy associated with exothermic nuclear reactions comes from the con-
version of mass into energy. If the mass loss is !M, then the energy released, Q ,
is given by Einstein’s relation, Q= (!M)c2, where c is the velocity of light. In this
section we discuss the energetics of nuclear transformations.
We first establish the quantitative relationship between atomic mass units
(AMU) and energy (MeV). By definition, the 12C atom has a mass of exactly
12 AMU. Since its gram atomic weight is 12 g, it follows that
1 AMU= 1/(6.02× 1023)= 1.66× 10–24 g= 1.66× 10–27 kg. (3.4)
Using the Einstein relation and c= 3× 108 m s–1, we obtain
1 AMU = (1.66× 10–27)(3× 108)2
= 1.49× 10–10 J (3.5)
= 1.49× 10
–10 J
1.6× 10–13 J MeV–1 = 931 MeV. (3.6)
More precisely, 1 AMU = 931.49 MeV.
We now consider one of the simplest nuclear reactions, the absorption of a ther-
mal neutron by a hydrogen atom, accompanied by emission of a gamma ray. This
reaction, which is very important for understanding the thermal-neutron dose to
the body, can be represented by writing
1
0n +
1
1H→ 21H + 00γ, (3.7)
the photon having zero charge and mass. The reaction can also be designated
1
1H(n,γ)
2
1H. To find the energy released, we compare the total masses on both sides
of the arrow. Appendix D contains data on nuclides which we shall frequently use.
The atomic weight M of a nuclide of mass number A can be found from the mass
difference, !, given in column 3. The quantity !=M – A gives the difference be-
tween the nuclide’s atomic weight and its atomic mass number, expressed in MeV.
(By definition, != 0 for the 12C atom.) Since we are interested only in energy dif-
ferences in the reaction (3.7), we obtain the energy released, Q, directly from the
values of !, without having to calculate the actual masses of the neutron and indi-
vidual atoms. Adding the ! values for 10n and
1
1H and subtracting that for
2
1H, we
find
Q= 8.0714 + 7.2890 – 13.1359= 2.2245 MeV. (3.8)
This energy appears as a gamma photon emitted when the capture takes place (the
thermal neutron has negligible kinetic energy).
3.2 Nuclear Binding Energies 59
greater than the energies associated with the valence electrons that are involved in
chemical reactions. This factor characterizes the enormous difference in the en-
ergy released when an atom undergoes a nuclear transformation as compared with
a chemical reaction.
The energy associated with exothermic nuclear reactions comes from the con-
version of mass into energy. If the mass loss is !M, then the energy released, Q ,
is given by Einstein’s relation, Q= (!M)c2, where c is the velocity of light. In this
section we discuss the energetics of nuclear transformations.
We first establish the quantitative relationship between atomic mass units
(AMU) and energy (MeV). By definition, the 12C atom has a mass of exactly
12 AMU. Since its gram atomic weight is 12 g, it follows that
1 AMU= 1/(6.02× 1023)= 1.66× 10–24 g= 1.66× 10–27 kg. (3.4)
Using the Einstein relation and c= 3× 108 m s–1, we obtain
1 AMU = (1.66× 10–27)(3× 108)2
= 1.49× 10–10 J (3.5)
= 1.49× 10
–10 J
1.6× 10–13 J MeV–1 = 931 MeV. (3.6)
More precisely, 1 AMU = 931.49 MeV.
We now consider one of the simplest nuclear reactions, the absorption of a ther-
mal neutron by a hydrogen atom, accompanied by emission of a gamma ray. This
reaction, which is very important for understanding the thermal-neutron dose to
the body, can be represented by writing
1
0n +
1
1H→ 21H + 00γ, (3.7)
the photon having zero charge and mass. The reaction can also be designated
1
1H(n,γ)
2
1H. To find the energy released, we compare the total masses on both sides
of the arrow. Appendix D contains data on nuclides which we shall frequently use.
The atomic weight M of a nuclide of mass number A can be found from themass
difference, !, given in column 3. The quantity !=M – A gives the difference be-
tween the nuclide’s atomic weight and its atomic mass number, expressed in MeV.
(By definition, != 0 for the 12C atom.) Since we are interested only in energy dif-
ferences in the reaction (3.7), we obtain the energy released, Q, directly from the
values of !, without having to calculate the actual masses of the neutron and indi-
vidual atoms. Adding the ! values for 10n and
1
1H and subtracting that for
2
1H, we
find
Q= 8.0714 + 7.2890 – 13.1359= 2.2245 MeV. (3.8)
This energy appears as a gamma photon emitted when the capture takes place (the
thermal neutron has negligible kinetic energy).
EXEMPLO
Determinar a energia do fóton liberado pela interação de um 
nêutron térmico com o hidrogênio. 
Considere a energia cinética do nêutron e hidrogênio iguais a zero. 
3.2 Nuclear Binding Energies 59
greater than the energies associated with the valence electrons that are involved in
chemical reactions. This factor characterizes the enormous difference in the en-
ergy released when an atom undergoes a nuclear transformation as compared with
a chemical reaction.
The energy associated with exothermic nuclear reactions comes from the con-
version of mass into energy. If the mass loss is !M, then the energy released, Q ,
is given by Einstein’s relation, Q= (!M)c2, where c is the velocity of light. In this
section we discuss the energetics of nuclear transformations.
We first establish the quantitative relationship between atomic mass units
(AMU) and energy (MeV). By definition, the 12C atom has a mass of exactly
12 AMU. Since its gram atomic weight is 12 g, it follows that
1 AMU= 1/(6.02× 1023)= 1.66× 10–24 g= 1.66× 10–27 kg. (3.4)
Using the Einstein relation and c= 3× 108 m s–1, we obtain
1 AMU = (1.66× 10–27)(3× 108)2
= 1.49× 10–10 J (3.5)
= 1.49× 10
–10 J
1.6× 10–13 J MeV–1 = 931 MeV. (3.6)
More precisely, 1 AMU = 931.49 MeV.
We now consider one of the simplest nuclear reactions, the absorption of a ther-
mal neutron by a hydrogen atom, accompanied by emission of a gamma ray. This
reaction, which is very important for understanding the thermal-neutron dose to
the body, can be represented by writing
1
0n +
1
1H→ 21H + 00γ, (3.7)
the photon having zero charge and mass. The reaction can also be designated
1
1H(n,γ)
2
1H. To find the energy released, we compare the total masses on both sides
of the arrow. Appendix D contains data on nuclides which we shall frequently use.
The atomic weight M of a nuclide of mass number A can be found from the mass
difference, !, given in column 3. The quantity !=M – A gives the difference be-
tween the nuclide’s atomic weight and its atomic mass number, expressed in MeV.
(By definition, != 0 for the 12C atom.) Since we are interested only in energy dif-
ferences in the reaction (3.7), we obtain the energy released, Q, directly from the
values of !, without having to calculate the actual masses of the neutron and indi-
vidual atoms. Adding the ! values for 10n and
1
1H and subtracting that for
2
1H, we
find
Q= 8.0714 + 7.2890 – 13.1359= 2.2245 MeV. (3.8)
This energy appears as a gamma photon emitted when the capture takes place (the
thermal neutron has negligible kinetic energy).
558
Appendix
D
.Selected
D
ata
on
N
uclides
Mass Difference Major Radiations,
Natural !=M –A (MeV) Energies (MeV), and
Abundance (at. mass – at. Type of Frequency per
Nuclide (%) mass No.) Decay Half-Life Disintegration (%)
1
0n — 8.0714 β
– 12 min β– : 0.78 max
1
1H 99.985 7.2890 — — —
2
1H 0.015 13.1359 — — —
3
1H — 14.9500 β
– 12.3 y β– : 0.0186 max, no γ
3
2He 0.00013 14.9313 — — —
4
2He 99.99+ 2.4248 — — —
6
3Li 7.42 14.088 — — —
7
3Li 92.58 14.907 — — —
7
4Be — 15.769 EC 53.3 d γ : 0.478 (10.3%)
10
5B 19.7 12.052 — — —
11
6C — 10.648 β
+ 99 +% 20.38 min β+ : 0.960 max (avg 0.386)
EC 0.2% γ : 0.511 (200%, γ±)
12
6C 98.892 0 — — —
14
6C — 3.0198 β
– 5730 y β– : 0.156 max (avg 0.0495), no γ
14
7N 99.635 2.8637 — — —
15
7N 0.356 0.100 — — —
16
8O 99.759 –4.7366 — — —
17
8O 0.037 –0.808 — — —
22
10Ne 8.82 –8.025 — — —
3.2 Nuclear Binding Energies 59
greater than the energies associated with the valence electrons that are involved in
chemical reactions. This factor characterizes the enormous difference in the en-
ergy released when an atom undergoes a nuclear transformation as compared with
a chemical reaction.
The energy associated with exothermic nuclear reactions comes from the con-
version of mass into energy. If the mass loss is !M, then the energy released, Q ,
is given by Einstein’s relation, Q= (!M)c2, where c is the velocity of light. In this
section we discuss the energetics of nuclear transformations.
We first establish the quantitative relationship between atomic mass units
(AMU) and energy (MeV). By definition, the 12C atom has a mass of exactly
12 AMU. Since its gram atomic weight is 12 g, it follows that
1 AMU= 1/(6.02× 1023)= 1.66× 10–24 g= 1.66× 10–27 kg. (3.4)
Using the Einstein relation and c= 3× 108 m s–1, we obtain
1 AMU = (1.66× 10–27)(3× 108)2
= 1.49× 10–10 J (3.5)
= 1.49× 10
–10 J
1.6× 10–13 J MeV–1 = 931 MeV. (3.6)
More precisely, 1 AMU = 931.49 MeV.
We now consider one of the simplest nuclear reactions, the absorption of a ther-
mal neutron by a hydrogen atom, accompanied by emission of a gamma ray. This
reaction, which is very important for understanding the thermal-neutron dose to
the body, can be represented by writing
1
0n +
1
1H→ 21H + 00γ, (3.7)
the photon having zero charge and mass. The reaction can also be designated
1
1H(n,γ)
2
1H. To find the energy released, we compare the total masses on both sides
of the arrow. Appendix D contains data on nuclides which we shall frequently use.
The atomic weight M of a nuclide of mass number A can be found from the mass
difference, !, given in column 3. The quantity !=M – A gives the difference be-
tween the nuclide’s atomic weight and its atomic mass number, expressed in MeV.
(By definition, != 0 for the 12C atom.) Since we are interested only in energy dif-
ferences in the reaction (3.7), we obtain the energy released, Q, directly from the
values of !, without having to calculate the actual masses of the neutron and indi-
vidual atoms. Adding the ! values for 10n and
1
1H and subtracting that for
2
1H, we
find
Q= 8.0714 + 7.2890 – 13.1359= 2.2245 MeV. (3.8)
This energy appears as a gamma photon emitted when the capture takes place (the
thermal neutron has negligible kinetic energy).
Massa Atômica - 
número de massa
(M - 1)*931 MeV
(M - 2)*931 MeV
3.2 Nuclear Binding Energies 59
greater than the energies associated with the valence electrons that are involved in
chemical reactions. This factor characterizes the enormous difference in the en-
ergy released when an atom undergoes a nuclear transformation as compared with
a chemical reaction.
The energy associated with exothermic nuclear reactions comes from the con-
version of mass into energy. If the mass loss is !M, then the energy released, Q ,
is given by Einstein’s relation, Q= (!M)c2, where c is the velocity of light. In this
section we discuss the energetics of nuclear transformations.
We first establish the quantitative relationship between atomic mass units
(AMU) and energy (MeV). By definition, the 12C atom has a mass of exactly
12 AMU. Since its gram atomic weight is 12 g, it follows that
1 AMU= 1/(6.02× 1023)= 1.66× 10–24 g= 1.66× 10–27 kg. (3.4)
Using the Einstein relation and c= 3× 108 m s–1, we obtain
1 AMU = (1.66× 10–27)(3× 108)2
= 1.49× 10–10 J (3.5)
= 1.49× 10
–10 J
1.6× 10–13 J MeV–1 = 931 MeV. (3.6)
More precisely, 1 AMU = 931.49 MeV.
We now consider one of the simplest nuclear reactions, the absorption ofa ther-
mal neutron by a hydrogen atom, accompanied by emission of a gamma ray. This
reaction, which is very important for understanding the thermal-neutron dose to
the body, can be represented by writing
1
0n +
1
1H→ 21H + 00γ, (3.7)
the photon having zero charge and mass. The reaction can also be designated
1
1H(n,γ)
2
1H. To find the energy released, we compare the total masses on both sides
of the arrow. Appendix D contains data on nuclides which we shall frequently use.
The atomic weight M of a nuclide of mass number A can be found from the mass
difference, !, given in column 3. The quantity !=M – A gives the difference be-
tween the nuclide’s atomic weight and its atomic mass number, expressed in MeV.
(By definition, != 0 for the 12C atom.) Since we are interested only in energy dif-
ferences in the reaction (3.7), we obtain the energy released, Q, directly from the
values of !, without having to calculate the actual masses of the neutron and indi-
vidual atoms. Adding the ! values for 10n and
1
1H and subtracting that for
2
1H, we
find
Q= 8.0714 + 7.2890 – 13.1359= 2.2245 MeV. (3.8)
This energy appears as a gamma photon emitted when the capture takes place (the
thermal neutron has negligible kinetic energy).
TIPOS DE RADIAÇÕES 
IONIZANTES
! Emissão Alfa 
! Emissão Isobárica (mesma massa atômica) 
! Emissão beta negativa 
! Emissão de pósitron 
! Captura eletrônica (captura de elétron orbital) 
! Emissão Isomérica (mesmas massa e no. atômico) 
! Gama 
! Conversão Interna
EMISSÃO ALFA
Rutherford E, Chadwick J, Ellis CD, Radiations from Radioactive 
Substances. NewYork, NY: Macmillan; 1930.
Radônio-220 Polônio-216
! Partícula alfa: 
! He++ (2 prótons+2 nêutrons) Mα = 4,00277 u.m.a. 
! Encontrado em elementos pesados, Z > 82 (exceto Sm-147 com 
Z=62)
EMISSÕES ALFA
+Po21084 X
?
?He
4
2
= α
EXEMPLO
O Ra-226 decai no Rn-222 emitindo uma partícula alfa.
64 3 The Nucleus and Nuclear Radiation
Fig. 3.4 Nuclear decay scheme of 22688Ra.
226Ra, since the excited state occurs in 5.5% of the total disintegrations and the
0.186 MeV photon is emitted only 3.3% of the time, it follows that internal con-
version occurs in about 2.2% of the total decays. As we show in more detail in
Section 3.6, the energy of the conversion electron is equal to the excited-state en-
ergy (in this case 0.186 MeV) minus the atomic-shell binding energy. The listing in
Appendix D shows one of the e– energies to be 0.170 MeV. In addition, since inter-
nal conversion leaves a K- or L-shell vacancy in the daughter atom, one also finds
among the photons emitted the characteristic X rays of Rn. Finally, as noted in the
radiations listed in Appendix D for 226Ra, various kinds of radiation are emitted
from the radioactive daughters, in this case 222Rn, 218Po, 214Pb, 214Bi, and 214Po.
Decay-scheme diagrams, such as that shown in Fig. 3.4 for 226Ra, conveniently
summarize the nuclear transformations. The two arrows slanting downward to the
left6) show the two modes of alpha decay along with the alpha-particle energies and
frequencies. Either changes the nucleus from that of 226Ra to that of 222Rn. When
the lower energy particle is emitted, the radon nucleus is left in an excited state with
energy 0.186 MeV above the ground state. (The vertical distances in Fig. 3.4 are
not to scale.) The subsequent gamma ray of this energy, which is emitted almost
immediately, is shown by the vertical wavy line. The frequency 3.3% associated
with this photon emission implies that an internal-conversion electron is emitted
in the other 2.2% of the total number of disintegrations. Radiations not emitted
directly from the nucleus (i.e., the Rn X rays and the internal-conversion electron)
are not shown on such a diagram, which represents the nuclear changes. Relatively
infrequent modes of decay could also be shown, but are not included in Fig. 3.4 (see
Fig. 3.7). (A small round-off error occurs in the energies.)
The most energetic alpha particles are found to come from radionuclides having
relatively short half-lives. An early empirical finding, known as the Geiger–Nuttall
law, implies that there is a linear relationship between the logarithm of the range R
6 By convention, going left represents a
decrease in Z and right, an increase in Z.
Photon emission is represented by a vertical
wavy line.
62 3 The Nucleus and Nuclear Radiation
3.3
Alpha Decay
Almost all naturally occurring alpha emitters are heavy elements with Z≥ 83. The
principal features of alpha decay can be learned from the example of 226Ra:
226
88Ra→ 22286Rn + 42He. (3.11)
The energy Q released in the decay arises from a net loss in the masses MRa,N,
MRn,N, andMHe,N, of the radium, radon, and helium nuclei:
Q=MRa,N –MRn,N –MHe,N. (3.12)
This nuclear mass difference is very nearly equal the atomic mass difference,
which, in turn, is equal to the difference in ! values.4) Letting !P, !D, and !He
denote the values of the parent, daughter, and helium atoms, we can write a general
equation for obtaining the energy release in alpha decay:
Qα =!P –!D –!He. (3.13)
Using the values in Appendix D for the decay of 22688Ra to the ground state of
222
86Rn,
we obtain
Q= 23.69 – 16.39 – 2.42= 4.88 MeV. (3.14)
The Q value (3.14) is shared by the alpha particle and the recoil radon nucleus,
and we can calculate the portion that each acquires. Since the radium nucleus was
at rest, the momenta of the two decay products must be equal and opposite. Letting
m and v represent the mass and initial velocity of the alpha particle and M and V
those of the recoil nucleus, we write
mv=MV. (3.15)
Since the initial kinetic energies of the products must be equal to the energy re-
leased in the decay, we have
1
2mv
2 + 12MV
2 =Q. (3.16)
Substituting V=mv/M from Eq. (3.15) into (3.16) and solving for v2, one finds
v2 = 2MQ
m(m +M)
. (3.17)
4 Specifically, the relatively slight difference in
the binding energies of the 88 electrons on
either side of the arrow in (3.11) is neglected
when atomic mass loss is equated to nuclear
mass loss. In principle, nuclear masses are
needed; however, atomic masses are much
better known. These small differences are
negligible for most purposes.
CONSERVAÇÃO DA MASSA
62 3 The Nucleus and Nuclear Radiation
3.3
Alpha Decay
Almost all naturally occurring alpha emitters are heavy elements with Z≥ 83. The
principal features of alpha decay can be learned from the example of 226Ra:
226
88Ra→ 22286Rn + 42He. (3.11)
The energy Q released in the decay arises from a net loss in the masses MRa,N,
MRn,N, andMHe,N, of the radium, radon, and helium nuclei:
Q=MRa,N –MRn,N –MHe,N. (3.12)
This nuclear mass difference is very nearly equal the atomic mass difference,
which, in turn, is equal to the difference in ! values.4) Letting !P, !D, and !He
denote the values of the parent, daughter, and helium atoms, we can write a general
equation for obtaining the energy release in alpha decay:
Qα =!P –!D –!He. (3.13)
Using the values in Appendix D for the decay of 22688Ra to the ground state of
222
86Rn,
we obtain
Q= 23.69 – 16.39 – 2.42= 4.88 MeV. (3.14)
The Q value (3.14) is shared by the alpha particle and the recoil radon nucleus,
and we can calculate the portion that each acquires. Since the radium nucleus was
at rest, the momenta of the two decay products must be equal and opposite. Letting
m and v represent the mass and initial velocity of the alpha particle and M and V
those of the recoil nucleus, we write
mv=MV. (3.15)
Since the initial kinetic energies of the products must be equal to the energy re-
leased in the decay, we have
1
2mv
2 + 12MV
2 =Q. (3.16)
Substituting V=mv/M from Eq. (3.15) into (3.16) and solving for v2, one finds
v2 = 2MQ
m(m +M)
. (3.17)
4 Specifically, therelatively slight difference in
the binding energies of the 88 electrons on
either side of the arrow in (3.11) is neglected
when atomic mass loss is equated to nuclear
mass loss. In principle, nuclear masses are
needed; however, atomic masses are much
better known. These small differences are
negligible for most purposes.
62 3 The Nucleus and Nuclear Radiation
3.3
Alpha Decay
Almost all naturally occurring alpha emitters are heavy elements with Z≥ 83. The
principal features of alpha decay can be learned from the example of 226Ra:
226
88Ra→ 22286Rn + 42He. (3.11)
The energy Q released in the decay arises from a net loss in the masses MRa,N,
MRn,N, andMHe,N, of the radium, radon, and helium nuclei:
Q=MRa,N –MRn,N –MHe,N. (3.12)
This nuclear mass difference is very nearly equal the atomic mass difference,
which, in turn, is equal to the difference in ! values.4) Letting !P, !D, and !He
denote the values of the parent, daughter, and helium atoms, we can write a general
equation for obtaining the energy release in alpha decay:
Qα =!P –!D –!He. (3.13)
Using the values in Appendix D for the decay of 22688Ra to the ground state of
222
86Rn,
we obtain
Q= 23.69 – 16.39 – 2.42= 4.88 MeV. (3.14)
The Q value (3.14) is shared by the alpha particle and the recoil radon nucleus,
and we can calculate the portion that each acquires. Since the radium nucleus was
at rest, the momenta of the two decay products must be equal and opposite. Letting
m and v represent the mass and initial velocity of the alpha particle and M and V
those of the recoil nucleus, we write
mv=MV. (3.15)
Since the initial kinetic energies of the products must be equal to the energy re-
leased in the decay, we have
1
2mv
2 + 12MV
2 =Q. (3.16)
Substituting V=mv/M from Eq. (3.15) into (3.16) and solving for v2, one finds
v2 = 2MQ
m(m +M)
. (3.17)
4 Specifically, the relatively slight difference in
the binding energies of the 88 electrons on
either side of the arrow in (3.11) is neglected
when atomic mass loss is equated to nuclear
mass loss. In principle, nuclear masses are
needed; however, atomic masses are much
better known. These small differences are
negligible for most purposes.
62 3 The Nucleus and Nuclear Radiation
3.3
Alpha Decay
Almost all naturally occurring alpha emitters are heavy elements with Z≥ 83. The
principal features of alpha decay can be learned from the example of 226Ra:
226
88Ra→ 22286Rn + 42He. (3.11)
The energy Q released in the decay arises from a net loss in the masses MRa,N,
MRn,N, andMHe,N, of the radium, radon, and helium nuclei:
Q=MRa,N –MRn,N –MHe,N. (3.12)
This nuclear mass difference is very nearly equal the atomic mass difference,
which, in turn, is equal to the difference in ! values.4) Letting !P, !D, and !He
denote the values of the parent, daughter, and helium atoms, we can write a general
equation for obtaining the energy release in alpha decay:
Qα =!P –!D –!He. (3.13)
Using the values in Appendix D for the decay of 22688Ra to the ground state of
222
86Rn,
we obtain
Q= 23.69 – 16.39 – 2.42= 4.88 MeV. (3.14)
The Q value (3.14) is shared by the alpha particle and the recoil radon nucleus,
and we can calculate the portion that each acquires. Since the radium nucleus was
at rest, the momenta of the two decay products must be equal and opposite. Letting
m and v represent the mass and initial velocity of the alpha particle and M and V
those of the recoil nucleus, we write
mv=MV. (3.15)
Since the initial kinetic energies of the products must be equal to the energy re-
leased in the decay, we have
1
2mv
2 + 12MV
2 =Q. (3.16)
Substituting V=mv/M from Eq. (3.15) into (3.16) and solving for v2, one finds
v2 = 2MQ
m(m +M)
. (3.17)
4 Specifically, the relatively slight difference in
the binding energies of the 88 electrons on
either side of the arrow in (3.11) is neglected
when atomic mass loss is equated to nuclear
mass loss. In principle, nuclear masses are
needed; however, atomic masses are much
better known. These small differences are
negligible for most purposes.
Ex:
566
Appendix
D
.Selected
D
ata
on
N
uclides
Mass Difference Major Radiations,
Natural !=M –A (MeV) Energies (MeV), and
Abundance (at. mass – at. Type of Frequency per
Nuclide (%) mass No.) Decay Half-Life Disintegration (%)
218
84Po 8.38 α (99.98%) 3.05 min α : 6.003
β–(0.02%) β– : 0.256 max
Daughter radiations from 214Pb, 214Bi,
214Po
219
86Rn — 8.85 α 3.96 s α : 6.82 (80%), 6.55 (12%),
6.42 (7%)
γ : 0.271 (10%), 0.402 (7%),
Po X rays
220
86Rn — 10.61 α 55.6 s α : 6.29
γ : 0.55 (0.07%)
Daughter radiations from 216Po
222
86Rn — 16.39 α 3.8235 d α : 5.490
γ : 0.510 (0.078%)
Daughter radiations from 218Po, 214Pb,
214Bi, 214Po
226
88Ra — 23.69 α 1600 y α : 4.602 (5.5%), 4.785 (94.4%)
γ : 0.186 (3.3%), 0.262 (0.005%),
Rn X rays
e– : 0.088, 0.170
Daughter radiations from 222Rn, 218Po,
214Pb, 214Bi, 214Po
235
92U 0.720 40.93 α 7.038× 108 y α : 4.218 (6%), 4.326 (5%), 4.366 (18%),
4.400 (56%), 4.417 (2%), 4.505
(2%), 4.558 (4%), 4.599 (5%)
558
Appendix
D
.Selected
D
ata
on
N
uclides
Mass Difference Major Radiations,
Natural !=M –A (MeV) Energies (MeV), and
Abundance (at. mass – at. Type of Frequency per
Nuclide (%) mass No.) Decay Half-Life Disintegration (%)
1
0n — 8.0714 β
– 12 min β– : 0.78 max
1
1H 99.985 7.2890 — — —
2
1H 0.015 13.1359 — — —
3
1H — 14.9500 β
– 12.3 y β– : 0.0186 max, no γ
3
2He 0.00013 14.9313 — — —
4
2He 99.99+ 2.4248 — — —
6
3Li 7.42 14.088 — — —
7
3Li 92.58 14.907 — — —
7
4Be — 15.769 EC 53.3 d γ : 0.478 (10.3%)
10
5B 19.7 12.052 — — —
11
6C — 10.648 β
+ 99 +% 20.38 min β+ : 0.960 max (avg 0.386)
EC 0.2% γ : 0.511 (200%, γ±)
12
6C 98.892 0 — — —
14
6C — 3.0198 β
– 5730 y β– : 0.156 max (avg 0.0495), no γ
14
7N 99.635 2.8637 — — —
15
7N 0.356 0.100 — — —
16
8O 99.759 –4.7366 — — —
17
8O 0.037 –0.808 — — —
22
10Ne 8.82 –8.025 — — —
CONSERVAÇÃO DA MASSA
62 3 The Nucleus and Nuclear Radiation
3.3
Alpha Decay
Almost all naturally occurring alpha emitters are heavy elements with Z≥ 83. The
principal features of alpha decay can be learned from the example of 226Ra:
226
88Ra→ 22286Rn + 42He. (3.11)
The energy Q released in the decay arises from a net loss in the masses MRa,N,
MRn,N, andMHe,N, of the radium, radon, and helium nuclei:
Q=MRa,N –MRn,N –MHe,N. (3.12)
This nuclear mass difference is very nearly equal the atomic mass difference,
which, in turn, is equal to the difference in ! values.4) Letting !P, !D, and !He
denote the values of the parent, daughter, and helium atoms, we can write a general
equation for obtaining the energy release in alpha decay:
Qα =!P –!D –!He. (3.13)
Using the values in Appendix D for the decay of 22688Ra to the ground state of
222
86Rn,
we obtain
Q= 23.69 – 16.39 – 2.42= 4.88 MeV. (3.14)
The Q value (3.14) is shared by the alpha particle and the recoil radon nucleus,
and we can calculate the portion that each acquires. Since the radium nucleus was
at rest, the momenta of the two decay products must be equal and opposite. Letting
m and v represent the mass and initial velocity of the alpha particle and M and V
those of the recoil nucleus, we write
mv=MV. (3.15)
Since the initial kinetic energies of the products must be equal to the energy re-
leased in the decay, we have
1
2mv
2 + 12MV
2 =Q. (3.16)
Substituting V=mv/M from Eq. (3.15) into (3.16) and solving for v2, one finds
v2 = 2MQ
m(m +M)
. (3.17)
4 Specifically, the relatively slight difference in
the binding energies of the 88 electrons on
either side of the arrow in (3.11) is neglected
when atomic mass loss is equated to nuclear
mass loss. In principle, nuclear masses are
needed; however, atomic massesare much
better known. These small differences are
negligible for most purposes.
62 3 The Nucleus and Nuclear Radiation
3.3
Alpha Decay
Almost all naturally occurring alpha emitters are heavy elements with Z≥ 83. The
principal features of alpha decay can be learned from the example of 226Ra:
226
88Ra→ 22286Rn + 42He. (3.11)
The energy Q released in the decay arises from a net loss in the masses MRa,N,
MRn,N, andMHe,N, of the radium, radon, and helium nuclei:
Q=MRa,N –MRn,N –MHe,N. (3.12)
This nuclear mass difference is very nearly equal the atomic mass difference,
which, in turn, is equal to the difference in ! values.4) Letting !P, !D, and !He
denote the values of the parent, daughter, and helium atoms, we can write a general
equation for obtaining the energy release in alpha decay:
Qα =!P –!D –!He. (3.13)
Using the values in Appendix D for the decay of 22688Ra to the ground state of
222
86Rn,
we obtain
Q= 23.69 – 16.39 – 2.42= 4.88 MeV. (3.14)
The Q value (3.14) is shared by the alpha particle and the recoil radon nucleus,
and we can calculate the portion that each acquires. Since the radium nucleus was
at rest, the momenta of the two decay products must be equal and opposite. Letting
m and v represent the mass and initial velocity of the alpha particle and M and V
those of the recoil nucleus, we write
mv=MV. (3.15)
Since the initial kinetic energies of the products must be equal to the energy re-
leased in the decay, we have
1
2mv
2 + 12MV
2 =Q. (3.16)
Substituting V=mv/M from Eq. (3.15) into (3.16) and solving for v2, one finds
v2 = 2MQ
m(m +M)
. (3.17)
4 Specifically, the relatively slight difference in
the binding energies of the 88 electrons on
either side of the arrow in (3.11) is neglected
when atomic mass loss is equated to nuclear
mass loss. In principle, nuclear masses are
needed; however, atomic masses are much
better known. These small differences are
negligible for most purposes.
62 3 The Nucleus and Nuclear Radiation
3.3
Alpha Decay
Almost all naturally occurring alpha emitters are heavy elements with Z≥ 83. The
principal features of alpha decay can be learned from the example of 226Ra:
226
88Ra→ 22286Rn + 42He. (3.11)
The energy Q released in the decay arises from a net loss in the masses MRa,N,
MRn,N, andMHe,N, of the radium, radon, and helium nuclei:
Q=MRa,N –MRn,N –MHe,N. (3.12)
This nuclear mass difference is very nearly equal the atomic mass difference,
which, in turn, is equal to the difference in ! values.4) Letting !P, !D, and !He
denote the values of the parent, daughter, and helium atoms, we can write a general
equation for obtaining the energy release in alpha decay:
Qα =!P –!D –!He. (3.13)
Using the values in Appendix D for the decay of 22688Ra to the ground state of
222
86Rn,
we obtain
Q= 23.69 – 16.39 – 2.42= 4.88 MeV. (3.14)
The Q value (3.14) is shared by the alpha particle and the recoil radon nucleus,
and we can calculate the portion that each acquires. Since the radium nucleus was
at rest, the momenta of the two decay products must be equal and opposite. Letting
m and v represent the mass and initial velocity of the alpha particle and M and V
those of the recoil nucleus, we write
mv=MV. (3.15)
Since the initial kinetic energies of the products must be equal to the energy re-
leased in the decay, we have
1
2mv
2 + 12MV
2 =Q. (3.16)
Substituting V=mv/M from Eq. (3.15) into (3.16) and solving for v2, one finds
v2 = 2MQ
m(m +M)
. (3.17)
4 Specifically, the relatively slight difference in
the binding energies of the 88 electrons on
either side of the arrow in (3.11) is neglected
when atomic mass loss is equated to nuclear
mass loss. In principle, nuclear masses are
needed; however, atomic masses are much
better known. These small differences are
negligible for most purposes.
Ex:
62 3 The Nucleus and Nuclear Radiation
3.3
Alpha Decay
Almost all naturally occurring alpha emitters are heavy elements with Z≥ 83. The
principal features of alpha decay can be learned from the example of 226Ra:
226
88Ra→ 22286Rn + 42He. (3.11)
The energy Q released in the decay arises from a net loss in the masses MRa,N,
MRn,N, andMHe,N, of the radium, radon, and helium nuclei:
Q=MRa,N –MRn,N –MHe,N. (3.12)
This nuclear mass difference is very nearly equal the atomic mass difference,
which, in turn, is equal to the difference in ! values.4) Letting !P, !D, and !He
denote the values of the parent, daughter, and helium atoms, we can write a general
equation for obtaining the energy release in alpha decay:
Qα =!P –!D –!He. (3.13)
Using the values in Appendix D for the decay of 22688Ra to the ground state of
222
86Rn,
we obtain
Q= 23.69 – 16.39 – 2.42= 4.88 MeV. (3.14)
The Q value (3.14) is shared by the alpha particle and the recoil radon nucleus,
and we can calculate the portion that each acquires. Since the radium nucleus was
at rest, the momenta of the two decay products must be equal and opposite. Letting
m and v represent the mass and initial velocity of the alpha particle and M and V
those of the recoil nucleus, we write
mv=MV. (3.15)
Since the initial kinetic energies of the products must be equal to the energy re-
leased in the decay, we have
1
2mv
2 + 12MV
2 =Q. (3.16)
Substituting V=mv/M from Eq. (3.15) into (3.16) and solving for v2, one finds
v2 = 2MQ
m(m +M)
. (3.17)
4 Specifically, the relatively slight difference in
the binding energies of the 88 electrons on
either side of the arrow in (3.11) is neglected
when atomic mass loss is equated to nuclear
mass loss. In principle, nuclear masses are
needed; however, atomic masses are much
better known. These small differences are
negligible for most purposes.
Unidades de Energia 
Joule J 
Elétron-volt 
1 eV = 1,6 10-19 J 
Unidades de Massa 
Kilograma kg 
Unidades de massa atômica 
1 uma = 1,660 . 10-27 kg
EXEMPLO
O Ra-226 decai no Rn-222 emitindo uma partícula alfa.
566
Appendix
D
.Selected
D
ata
on
N
uclides
Mass Difference Major Radiations,
Natural !=M –A (MeV) Energies (MeV), and
Abundance (at. mass – at. Type of Frequency per
Nuclide (%) mass No.) Decay Half-Life Disintegration (%)
218
84Po 8.38 α (99.98%) 3.05 min α : 6.003
β–(0.02%) β– : 0.256 max
Daughter radiations from 214Pb, 214Bi,
214Po
219
86Rn — 8.85 α 3.96 s α : 6.82 (80%), 6.55 (12%),
6.42 (7%)
γ : 0.271 (10%), 0.402 (7%),
Po X rays
220
86Rn — 10.61 α 55.6 s α : 6.29
γ : 0.55 (0.07%)
Daughter radiations from 216Po
222
86Rn — 16.39 α 3.8235 d α : 5.490
γ : 0.510 (0.078%)
Daughter radiations from 218Po, 214Pb,
214Bi, 214Po
226
88Ra — 23.69 α 1600 y α : 4.602 (5.5%), 4.785 (94.4%)
γ : 0.186 (3.3%), 0.262 (0.005%),
Rn X rays
e– : 0.088, 0.170
Daughter radiations from 222Rn, 218Po,
214Pb, 214Bi, 214Po
235
92U 0.720 40.93 α 7.038× 108 y α : 4.218 (6%), 4.326 (5%), 4.366 (18%),
4.400 (56%), 4.417 (2%), 4.505
(2%), 4.558 (4%), 4.599 (5%)
64 3 The Nucleus and Nuclear Radiation
Fig. 3.4 Nuclear decay scheme of 22688Ra.
226Ra, since the excited state occurs in 5.5% of the total disintegrations and the
0.186 MeV photon is emitted only 3.3% of the time, it follows that internal con-
version occurs in about 2.2% of the total decays. As we show in more detail in
Section 3.6, the energy of the conversion electron is equal to the excited-state en-
ergy (in this case 0.186 MeV) minus the atomic-shell binding energy. The listing in
Appendix D shows one of the e– energies to be 0.170 MeV. In addition, since inter-
nal conversion leaves a K- or L-shell vacancy in the daughter atom, one also finds
among the photons emitted the characteristic X rays of Rn. Finally, as noted in the
radiations listed in Appendix D for 226Ra, various kinds of radiation are emitted
from the radioactive daughters, in this case 222Rn,218Po, 214Pb, 214Bi, and 214Po.
Decay-scheme diagrams, such as that shown in Fig. 3.4 for 226Ra, conveniently
summarize the nuclear transformations. The two arrows slanting downward to the
left6) show the two modes of alpha decay along with the alpha-particle energies and
frequencies. Either changes the nucleus from that of 226Ra to that of 222Rn. When
the lower energy particle is emitted, the radon nucleus is left in an excited state with
energy 0.186 MeV above the ground state. (The vertical distances in Fig. 3.4 are
not to scale.) The subsequent gamma ray of this energy, which is emitted almost
immediately, is shown by the vertical wavy line. The frequency 3.3% associated
with this photon emission implies that an internal-conversion electron is emitted
in the other 2.2% of the total number of disintegrations. Radiations not emitted
directly from the nucleus (i.e., the Rn X rays and the internal-conversion electron)
are not shown on such a diagram, which represents the nuclear changes. Relatively
infrequent modes of decay could also be shown, but are not included in Fig. 3.4 (see
Fig. 3.7). (A small round-off error occurs in the energies.)
The most energetic alpha particles are found to come from radionuclides having
relatively short half-lives. An early empirical finding, known as the Geiger–Nuttall
law, implies that there is a linear relationship between the logarithm of the range R
6 By convention, going left represents a
decrease in Z and right, an increase in Z.
Photon emission is represented by a vertical
wavy line.
62 3 The Nucleus and Nuclear Radiation
3.3
Alpha Decay
Almost all naturally occurring alpha emitters are heavy elements with Z≥ 83. The
principal features of alpha decay can be learned from the example of 226Ra:
226
88Ra→ 22286Rn + 42He. (3.11)
The energy Q released in the decay arises from a net loss in the masses MRa,N,
MRn,N, andMHe,N, of the radium, radon, and helium nuclei:
Q=MRa,N –MRn,N –MHe,N. (3.12)
This nuclear mass difference is very nearly equal the atomic mass difference,
which, in turn, is equal to the difference in ! values.4) Letting !P, !D, and !He
denote the values of the parent, daughter, and helium atoms, we can write a general
equation for obtaining the energy release in alpha decay:
Qα =!P –!D –!He. (3.13)
Using the values in Appendix D for the decay of 22688Ra to the ground state of
222
86Rn,
we obtain
Q= 23.69 – 16.39 – 2.42= 4.88 MeV. (3.14)
The Q value (3.14) is shared by the alpha particle and the recoil radon nucleus,
and we can calculate the portion that each acquires. Since the radium nucleus was
at rest, the momenta of the two decay products must be equal and opposite. Letting
m and v represent the mass and initial velocity of the alpha particle and M and V
those of the recoil nucleus, we write
mv=MV. (3.15)
Since the initial kinetic energies of the products must be equal to the energy re-
leased in the decay, we have
1
2mv
2 + 12MV
2 =Q. (3.16)
Substituting V=mv/M from Eq. (3.15) into (3.16) and solving for v2, one finds
v2 = 2MQ
m(m +M)
. (3.17)
4 Specifically, the relatively slight difference in
the binding energies of the 88 electrons on
either side of the arrow in (3.11) is neglected
when atomic mass loss is equated to nuclear
mass loss. In principle, nuclear masses are
needed; however, atomic masses are much
better known. These small differences are
negligible for most purposes.
62 3 The Nucleus and Nuclear Radiation
3.3
Alpha Decay
Almost all naturally occurring alpha emitters are heavy elements with Z≥ 83. The
principal features of alpha decay can be learned from the example of 226Ra:
226
88Ra→ 22286Rn + 42He. (3.11)
The energy Q released in the decay arises from a net loss in the masses MRa,N,
MRn,N, andMHe,N, of the radium, radon, and helium nuclei:
Q=MRa,N –MRn,N –MHe,N. (3.12)
This nuclear mass difference is very nearly equal the atomic mass difference,
which, in turn, is equal to the difference in ! values.4) Letting !P, !D, and !He
denote the values of the parent, daughter, and helium atoms, we can write a general
equation for obtaining the energy release in alpha decay:
Qα =!P –!D –!He. (3.13)
Using the values in Appendix D for the decay of 22688Ra to the ground state of
222
86Rn,
we obtain
Q= 23.69 – 16.39 – 2.42= 4.88 MeV. (3.14)
The Q value (3.14) is shared by the alpha particle and the recoil radon nucleus,
and we can calculate the portion that each acquires. Since the radium nucleus was
at rest, the momenta of the two decay products must be equal and opposite. Letting
m and v represent the mass and initial velocity of the alpha particle and M and V
those of the recoil nucleus, we write
mv=MV. (3.15)
Since the initial kinetic energies of the products must be equal to the energy re-
leased in the decay, we have
1
2mv
2 + 12MV
2 =Q. (3.16)
Substituting V=mv/M from Eq. (3.15) into (3.16) and solving for v2, one finds
v2 = 2MQ
m(m +M)
. (3.17)
4 Specifically, the relatively slight difference in
the binding energies of the 88 electrons on
either side of the arrow in (3.11) is neglected
when atomic mass loss is equated to nuclear
mass loss. In principle, nuclear masses are
needed; however, atomic masses are much
better known. These small differences are
negligible for most purposes.
62 3 The Nucleus and Nuclear Radiation
3.3
Alpha Decay
Almost all naturally occurring alpha emitters are heavy elements with Z≥ 83. The
principal features of alpha decay can be learned from the example of 226Ra:
226
88Ra→ 22286Rn + 42He. (3.11)
The energy Q released in the decay arises from a net loss in the masses MRa,N,
MRn,N, andMHe,N, of the radium, radon, and helium nuclei:
Q=MRa,N –MRn,N –MHe,N. (3.12)
This nuclear mass difference is very nearly equal the atomic mass difference,
which, in turn, is equal to the difference in ! values.4) Letting !P, !D, and !He
denote the values of the parent, daughter, and helium atoms, we can write a general
equation for obtaining the energy release in alpha decay:
Qα =!P –!D –!He. (3.13)
Using the values in Appendix D for the decay of 22688Ra to the ground state of
222
86Rn,
we obtain
Q= 23.69 – 16.39 – 2.42= 4.88 MeV. (3.14)
The Q value (3.14) is shared by the alpha particle and the recoil radon nucleus,
and we can calculate the portion that each acquires. Since the radium nucleus was
at rest, the momenta of the two decay products must be equal and opposite. Letting
m and v represent the mass and initial velocity of the alpha particle and M and V
those of the recoil nucleus, we write
mv=MV. (3.15)
Since the initial kinetic energies of the products must be equal to the energy re-
leased in the decay, we have
1
2mv
2 + 12MV
2 =Q. (3.16)
Substituting V=mv/M from Eq. (3.15) into (3.16) and solving for v2, one finds
v2 = 2MQ
m(m +M)
. (3.17)
4 Specifically, the relatively slight difference in
the binding energies of the 88 electrons on
either side of the arrow in (3.11) is neglected
when atomic mass loss is equated to nuclear
mass loss. In principle, nuclear masses are
needed; however, atomic masses are much
better known. These small differences are
negligible for most purposes.
62 3 The Nucleus and Nuclear Radiation
3.3
Alpha Decay
Almost all naturally occurring alpha emitters are heavy elements with Z≥ 83. The
principal features of alpha decay can be learned from the example of 226Ra:
226
88Ra→ 22286Rn + 42He. (3.11)
The energy Q released in the decay arises from a net loss in the masses MRa,N,
MRn,N, andMHe,N, of the radium, radon, and helium nuclei:
Q=MRa,N –MRn,N –MHe,N. (3.12)
This nuclear mass difference is very nearly equal the atomic mass difference,
which, in turn, is equal to the difference in ! values.4) Letting !P, !D, and !He
denote the values of the parent, daughter, and helium atoms, we can writea general
equation for obtaining the energy release in alpha decay:
Qα =!P –!D –!He. (3.13)
Using the values in Appendix D for the decay of 22688Ra to the ground state of
222
86Rn,
we obtain
Q= 23.69 – 16.39 – 2.42= 4.88 MeV. (3.14)
The Q value (3.14) is shared by the alpha particle and the recoil radon nucleus,
and we can calculate the portion that each acquires. Since the radium nucleus was
at rest, the momenta of the two decay products must be equal and opposite. Letting
m and v represent the mass and initial velocity of the alpha particle and M and V
those of the recoil nucleus, we write
mv=MV. (3.15)
Since the initial kinetic energies of the products must be equal to the energy re-
leased in the decay, we have
1
2mv
2 + 12MV
2 =Q. (3.16)
Substituting V=mv/M from Eq. (3.15) into (3.16) and solving for v2, one finds
v2 = 2MQ
m(m +M)
. (3.17)
4 Specifically, the relatively slight difference in
the binding energies of the 88 electrons on
either side of the arrow in (3.11) is neglected
when atomic mass loss is equated to nuclear
mass loss. In principle, nuclear masses are
needed; however, atomic masses are much
better known. These small differences are
negligible for most purposes.
62 3 The Nucleus and Nuclear Radiation
3.3
Alpha Decay
Almost all naturally occurring alpha emitters are heavy elements with Z≥ 83. The
principal features of alpha decay can be learned from the example of 226Ra:
226
88Ra→ 22286Rn + 42He. (3.11)
The energy Q released in the decay arises from a net loss in the masses MRa,N,
MRn,N, andMHe,N, of the radium, radon, and helium nuclei:
Q=MRa,N –MRn,N –MHe,N. (3.12)
This nuclear mass difference is very nearly equal the atomic mass difference,
which, in turn, is equal to the difference in ! values.4) Letting !P, !D, and !He
denote the values of the parent, daughter, and helium atoms, we can write a general
equation for obtaining the energy release in alpha decay:
Qα =!P –!D –!He. (3.13)
Using the values in Appendix D for the decay of 22688Ra to the ground state of
222
86Rn,
we obtain
Q= 23.69 – 16.39 – 2.42= 4.88 MeV. (3.14)
The Q value (3.14) is shared by the alpha particle and the recoil radon nucleus,
and we can calculate the portion that each acquires. Since the radium nucleus was
at rest, the momenta of the two decay products must be equal and opposite. Letting
m and v represent the mass and initial velocity of the alpha particle and M and V
those of the recoil nucleus, we write
mv=MV. (3.15)
Since the initial kinetic energies of the products must be equal to the energy re-
leased in the decay, we have
1
2mv
2 + 12MV
2 =Q. (3.16)
Substituting V=mv/M from Eq. (3.15) into (3.16) and solving for v2, one finds
v2 = 2MQ
m(m +M)
. (3.17)
4 Specifically, the relatively slight difference in
the binding energies of the 88 electrons on
either side of the arrow in (3.11) is neglected
when atomic mass loss is equated to nuclear
mass loss. In principle, nuclear masses are
needed; however, atomic masses are much
better known. These small differences are
negligible for most purposes.
62 3 The Nucleus and Nuclear Radiation
3.3
Alpha Decay
Almost all naturally occurring alpha emitters are heavy elements with Z≥ 83. The
principal features of alpha decay can be learned from the example of 226Ra:
226
88Ra→ 22286Rn + 42He. (3.11)
The energy Q released in the decay arises from a net loss in the masses MRa,N,
MRn,N, andMHe,N, of the radium, radon, and helium nuclei:
Q=MRa,N –MRn,N –MHe,N. (3.12)
This nuclear mass difference is very nearly equal the atomic mass difference,
which, in turn, is equal to the difference in ! values.4) Letting !P, !D, and !He
denote the values of the parent, daughter, and helium atoms, we can write a general
equation for obtaining the energy release in alpha decay:
Qα =!P –!D –!He. (3.13)
Using the values in Appendix D for the decay of 22688Ra to the ground state of
222
86Rn,
we obtain
Q= 23.69 – 16.39 – 2.42= 4.88 MeV. (3.14)
The Q value (3.14) is shared by the alpha particle and the recoil radon nucleus,
and we can calculate the portion that each acquires. Since the radium nucleus was
at rest, the momenta of the two decay products must be equal and opposite. Letting
m and v represent the mass and initial velocity of the alpha particle and M and V
those of the recoil nucleus, we write
mv=MV. (3.15)
Since the initial kinetic energies of the products must be equal to the energy re-
leased in the decay, we have
1
2mv
2 + 12MV
2 =Q. (3.16)
Substituting V=mv/M from Eq. (3.15) into (3.16) and solving for v2, one finds
v2 = 2MQ
m(m +M)
. (3.17)
4 Specifically, the relatively slight difference in
the binding energies of the 88 electrons on
either side of the arrow in (3.11) is neglected
when atomic mass loss is equated to nuclear
mass loss. In principle, nuclear masses are
needed; however, atomic masses are much
better known. These small differences are
negligible for most purposes.
62 3 The Nucleus and Nuclear Radiation
3.3
Alpha Decay
Almost all naturally occurring alpha emitters are heavy elements with Z≥ 83. The
principal features of alpha decay can be learned from the example of 226Ra:
226
88Ra→ 22286Rn + 42He. (3.11)
The energy Q released in the decay arises from a net loss in the masses MRa,N,
MRn,N, andMHe,N, of the radium, radon, and helium nuclei:
Q=MRa,N –MRn,N –MHe,N. (3.12)
This nuclear mass difference is very nearly equal the atomic mass difference,
which, in turn, is equal to the difference in ! values.4) Letting !P, !D, and !He
denote the values of the parent, daughter, and helium atoms, we can write a general
equation for obtaining the energy release in alpha decay:
Qα =!P –!D –!He. (3.13)
Using the values in Appendix D for the decay of 22688Ra to the ground state of
222
86Rn,
we obtain
Q= 23.69 – 16.39 – 2.42= 4.88 MeV. (3.14)
The Q value (3.14) is shared by the alpha particle and the recoil radon nucleus,
and we can calculate the portion that each acquires. Since the radium nucleus was
at rest, the momenta of the two decay products must be equal and opposite. Letting
m and v represent the mass and initial velocity of the alpha particle and M and V
those of the recoil nucleus, we write
mv=MV. (3.15)
Since the initial kinetic energies of the products must be equal to the energy re-
leased in the decay, we have
1
2mv
2 + 12MV
2 =Q. (3.16)
Substituting V=mv/M from Eq. (3.15) into (3.16) and solving for v2, one finds
v2 = 2MQ
m(m +M)
. (3.17)
4 Specifically, the relatively slight difference in
the binding energies of the 88 electrons on
either side of the arrow in (3.11) is neglected
when atomic mass loss is equated to nuclear
mass loss. In principle, nuclear masses are
needed; however, atomic masses are much
better known. These small differences are
negligible for most purposes.
3.3 Alpha Decay 63
One thus obtains for the alpha-particle energy
Eα = 12mv2 =
MQ
m +M
. (3.18)
With the roles of the two masses interchanged, it follows that the recoil energy of
the nucleus is
EN = 12MV2 =
mQ
m +M
. (3.19)
As a check, we see that Eα + EN =Q. Because of its much smaller mass, the alpha
particle, having the same momentum as the nucleus, has much more energy. For
226Ra, it follows from (3.14) and (3.18) that
Eα = 222× 4.884 + 222 = 4.79 MeV. (3.20)
The radon nucleus recoils with an energy of only 0.09 MeV.
The conservation of momentum and energy, Eqs. (3.15) and (3.16), fixes the en-
ergy of an alpha particle uniquely for given values of Q and M. Alpha particles
therefore occur with discrete values of energy.
Appendix D gives the principal radiations emitted by various nuclides. We
consider each of those listed for 226Ra. Two alpha-particle energies are shown:
4.785 MeV, occurring with a frequency of 94.4% ofall decays, and 4.602 MeV, oc-
curring 5.5% of the time. The Q value for the less frequent alpha particle can be
found from Eq. (3.18):
Q= (m +M)Eα
M
= 226× 4.60
222
= 4.68 MeV. (3.21)
The decay in this case goes to an excited state of the 222Rn nucleus. Like excited
atomic states, excited nuclear states can decay by photon emission. Photons from
the nucleus are called gamma rays, and their energies are generally in the range
from tens of keV to several MeV. Under the gamma rays listed in Appendix D for
226Ra we find a 0.186-MeV photon emitted in 3.3% of the decays, in addition to
another that occurs very infrequently (following alpha decay to still another excited
level of higher energy in the daughter nucleus). We conclude that emission of the
higher energy alpha particle (Eα = 4.79 MeV) leaves the daughter 222Rn nucleus in
its ground state. Emission of the 4.60 MeV alpha particle leaves the nucleus in an
excited state with energy 4.79 – 4.60= 0.19 MeV above the ground state. A photon
of this energy can then be emitted from the nucleus, and, indeed, one of energy
0.186MeV is listed for 3.3% of the decays. As an alternative to photon emission, un-
der certain circumstances an excited nuclear state can decay by ejecting an atomic
electron, usually from the K or L shell. This process, which produces the electrons
listed (e–), is called internal conversion, and will be discussed in Section 3.6.5) For
5 In atoms, an Auger electron can be ejected
from a shell in place of a photon,
accompanying an electronic transition (Sect.
2.11).
3.3 Alpha Decay 63
One thus obtains for the alpha-particle energy
Eα = 12mv2 =
MQ
m +M
. (3.18)
With the roles of the two masses interchanged, it follows that the recoil energy of
the nucleus is
EN = 12MV2 =
mQ
m +M
. (3.19)
As a check, we see that Eα + EN =Q. Because of its much smaller mass, the alpha
particle, having the same momentum as the nucleus, has much more energy. For
226Ra, it follows from (3.14) and (3.18) that
Eα = 222× 4.884 + 222 = 4.79 MeV. (3.20)
The radon nucleus recoils with an energy of only 0.09 MeV.
The conservation of momentum and energy, Eqs. (3.15) and (3.16), fixes the en-
ergy of an alpha particle uniquely for given values of Q and M. Alpha particles
therefore occur with discrete values of energy.
Appendix D gives the principal radiations emitted by various nuclides. We
consider each of those listed for 226Ra. Two alpha-particle energies are shown:
4.785 MeV, occurring with a frequency of 94.4% of all decays, and 4.602 MeV, oc-
curring 5.5% of the time. The Q value for the less frequent alpha particle can be
found from Eq. (3.18):
Q= (m +M)Eα
M
= 226× 4.60
222
= 4.68 MeV. (3.21)
The decay in this case goes to an excited state of the 222Rn nucleus. Like excited
atomic states, excited nuclear states can decay by photon emission. Photons from
the nucleus are called gamma rays, and their energies are generally in the range
from tens of keV to several MeV. Under the gamma rays listed in Appendix D for
226Ra we find a 0.186-MeV photon emitted in 3.3% of the decays, in addition to
another that occurs very infrequently (following alpha decay to still another excited
level of higher energy in the daughter nucleus). We conclude that emission of the
higher energy alpha particle (Eα = 4.79 MeV) leaves the daughter 222Rn nucleus in
its ground state. Emission of the 4.60 MeV alpha particle leaves the nucleus in an
excited state with energy 4.79 – 4.60= 0.19 MeV above the ground state. A photon
of this energy can then be emitted from the nucleus, and, indeed, one of energy
0.186MeV is listed for 3.3% of the decays. As an alternative to photon emission, un-
der certain circumstances an excited nuclear state can decay by ejecting an atomic
electron, usually from the K or L shell. This process, which produces the electrons
listed (e–), is called internal conversion, and will be discussed in Section 3.6.5) For
5 In atoms, an Auger electron can be ejected
from a shell in place of a photon,
accompanying an electronic transition (Sect.
2.11).
3.3 Alpha Decay 63
One thus obtains for the alpha-particle energy
Eα = 12mv2 =
MQ
m +M
. (3.18)
With the roles of the two masses interchanged, it follows that the recoil energy of
the nucleus is
EN = 12MV2 =
mQ
m +M
. (3.19)
As a check, we see that Eα + EN =Q. Because of its much smaller mass, the alpha
particle, having the same momentum as the nucleus, has much more energy. For
226Ra, it follows from (3.14) and (3.18) that
Eα = 222× 4.884 + 222 = 4.79 MeV. (3.20)
The radon nucleus recoils with an energy of only 0.09 MeV.
The conservation of momentum and energy, Eqs. (3.15) and (3.16), fixes the en-
ergy of an alpha particle uniquely for given values of Q and M. Alpha particles
therefore occur with discrete values of energy.
Appendix D gives the principal radiations emitted by various nuclides. We
consider each of those listed for 226Ra. Two alpha-particle energies are shown:
4.785 MeV, occurring with a frequency of 94.4% of all decays, and 4.602 MeV, oc-
curring 5.5% of the time. The Q value for the less frequent alpha particle can be
found from Eq. (3.18):
Q= (m +M)Eα
M
= 226× 4.60
222
= 4.68 MeV. (3.21)
The decay in this case goes to an excited state of the 222Rn nucleus. Like excited
atomic states, excited nuclear states can decay by photon emission. Photons from
the nucleus are called gamma rays, and their energies are generally in the range
from tens of keV to several MeV. Under the gamma rays listed in Appendix D for
226Ra we find a 0.186-MeV photon emitted in 3.3% of the decays, in addition to
another that occurs very infrequently (following alpha decay to still another excited
level of higher energy in the daughter nucleus). We conclude that emission of the
higher energy alpha particle (Eα = 4.79 MeV) leaves the daughter 222Rn nucleus in
its ground state. Emission of the 4.60 MeV alpha particle leaves the nucleus in an
excited state with energy 4.79 – 4.60= 0.19 MeV above the ground state. A photon
of this energy can then be emitted from the nucleus, and, indeed, one of energy
0.186MeV is listed for 3.3% of the decays. As an alternative to photon emission, un-
der certain circumstances an excited nuclear state can decay by ejecting an atomic
electron, usually from the K or L shell. This process, which produces the electrons
listed (e–), is called internal conversion, and will be discussed in Section 3.6.5) For
5 In atoms, an Auger electron can be ejected
from a shell in place of a photon,
accompanying an electronic transition (Sect.
2.11).
Partículas alfa (ou beta) 
Alcance de 40 to 90 µm 
2 a 10 células 
Bi211, Bi212, Bi213
RADIO-IMMUNOTHERAPY 
OU TARGETED ALPHA THERAPY
http://bloodjournal.hematologylibrary.org/cgi/content/full/100/4/1233
Fonte: D. Scott Wilbur, Ph.D., Department of Radiation Oncology, University of Washington
DISTRIBUTION OF [211-AT]ASTATIDE AND [131-I]IODIDE IN RATS 
(SELECTED TISSUES) 
Fonte: D. Scott Wilbur, Ph.D., Department of Radiation Oncology, University of Washington
The smoke detector has two ionization chambers, 
one open to the air, and a reference chamber which 
does not allow the entry of particles. 
The radioactive source emits alpha particles into 
both chambers, which ionizes some air molecules. 
There is a potential difference (voltage) between 
pairs of electrodes in the chambers; the electrical 
charge on the ions allows an electric current to flow.
 
If any smoke particles enter the open chamber, 
some of the ions will attach to the particles and not 
be available to carry the current in that chamber. An 
electronic circuit detects that a current difference 
has developed between the open and sealed 
chambers, and sounds the alarm
Smoke detector
Staticmaster ®
In-Line Alpha Ionizers
2937 Alt Boulevard/ PO Box 310 / Grand Island, NY 14072-0310
TOLL FREE 800 525 8076 / PHONE 716 773 7634 / FAX 716 773 7744 / EMAIL sales@nrdinc.com / WEB www.nrdinc.com
The model P-2021 is a simple, reliable, 
efficient and effective means of eliminating
static in a variety of industries, including 
electronics, optics, graphics and finishing. 
Easy to install on any clean, dry compressed air
or inert gas line, the P-2021 emits a powerful
stream of ions to quickly neutralize electrostatic
charges. By design, the P-2021 never goes out 
of balance, so it does not require periodic 
calibration. In addition, it requires no electrical
power to operate, so it’s intrinsically safe for 
use in flammable or explosive environments. 
The model P-2031 is designed to deliver
high volumes of ionized compressed gases at
low pressure. Similar to the P-2021, it is easily
installed on any clean, dry compressed air or
inert gas line. As the compressed gas passes
through the device, it is ionized and does not
require a specific distance from the affected 
surface to achieve proper ion intermix. The 
P-2031 always produces a balanced ion output
down to zero volts. The P-2031 is ideally suited
for removing static charges in storage containers
and hoppers, such as those used in the recycling
of plastic scrap or pelletizers where a large 
area of coverage is required. 
Staticmaster ®
In-Line Alpha Ionizers
Notice: NRD warrants that these materials are reasonably fit for the purpose described. We believe that specifications and technical information
presented are based on test methods that produce accurate results. The foregoing limited warranty is expressly made in lieu of, and no other
express and no implied warranty of merchantability or fitness for a particular purpose is made. NRD, LLC, will not be liable for incidental, 
consequential, special, exemplary or punitive damages. The limited warranty will not apply to defects or damage due to accidents, neglect, 
misuse, alterations, error or failure to properly maintain, clean or repair products.
2937 Alt Boulevard / PO Box 310 / Grand Island, NY 14072-0310
TOLL FREE 800 525 8076 / PHONE 716 773 7634 / FAX 716 773 7744 / EMAIL sales@nrdinc.com / WEB www.nrdinc.com
Note on Ion Source
All NRD ion sources are encapsulated in precious metals by means of a patented process. This results in a static eliminator which resists 
oxidation, solvents, heat, cold and vibration.
To maintain peak performance and conform to government regulations, ionizers are available on a one-year lease. 
Model P-2021 In-Line Ionizer
Specifications
Diameter: 0.50" (12.7 mm)
Weight: 1.3 oz (37 g)
Material: Stainless steel
Gas: Clean, dry compressed air or inert gas
Pressure: Normal: <40 psig (2.8 bar)
Maximum: 100 psig (6.9 bar)
Temperature: Maximum temperature 310° F 
(155° C)
Isotope: Polonium-210
Emission: Alpha
Activity: 10 mCi (370 MBq)
Line Connections: 1/8" – 27 NPSM thread
male/female
Decay Time: 0.7 seconds (ESD Association 3.1 
@ 30 psig, 12" distance, 1,000 to 100 V)
Model P-2031 Specifications
Diameter: 1" (25.4 mm)
Weight: 3.5 oz (100 g)/10 oz (286 g)
Material: Aluminum/316 stainless steel
Gas: Clean, dry compressed air or inert gas
Pressure: Normal: <20 psig (1.4 bar)
Maximum: 100 psig (6.9 bar)
Isotope: Polonium-210
Emission: Alpha
Activity: 20 mCi (740 MBq)
Line Connections: 1/4" NPT male and female
Decay Time: 0.5 seconds (ESD Association 3.1 
@ 10 psig, 12" distance, 1,000 to 100 V)
Ionization Efficiency
Air Supply Pressure PSIG
O
pt
im
al
 P
ea
k
Io
ni
za
ti
on
 E
ffi
ci
en
cy
Staticmaster ®
Microbalance Ionizers and Brushes
2937 Alt Boulevard / PO Box 310 / Grand Island, NY 14072-0310
TOLL FREE 800 525 8076 / PHONE 716 773 7634 / FAX 716 773 7744 / EMAIL sales@nrdinc.com / WEB www.nrdinc.com
When weighing fine powders and fibers, the
balance chamber must maintain a stable static
free environment to assure accurate results. 
The problem of “fly-away” powders and 
inaccurate weighing is eliminated with the
2U500 and 1U400 microbalance 
ionizers, because they utilize proven alpha
ionizing technology. These devices can be easily
mounted within the balance chamber, assuring
quick and accurate measurements. 
Static isn’t a problem when weighing filters
with the Duostat™ BF2-1000 because 
it completely eliminates any charge on both
sides of the filter. The result is increased 
accuracy and efficiency for reliable and 
repeatable weighing.
Also utilizing alpha ionizing technology to 
eliminate static in a wide variety of industrial,
photographic and medical applications are 
the 3C500 and 1C200 static 
neutralizing brushes.
These ionizing brushes are hand-crafted with
soft, natural hair that is safe to use with even
the most delicate lens coatings, film emulsions
and magnetic media. 
Note on Ion Source
All NRD ion sources are encapsulated in precious metals by means of a 
patented process. This results in a static eliminator which resists oxidation, 
solvents, heat, cold and vibration.
1C200 1" brush3C500 3" brush
BF2- 1000 positioner
2U500 microbalance ionizer
1U400 microbalance ionizer
Notice: NRD warrants that these materials are reasonably fit for the purpose described. We believe that specifications and technical information
presented are based on test methods that produce accurate results. The foregoing limited warranty is expressly made in lieu of, and no other
express and no implied warranty of merchantability or fitness for a particular purpose is made. NRD, LLC, will not be liable for incidental, 
consequential, special, exemplary or punitive damages. The limited warranty will not apply to defects or damage due to accidents, neglect, 
misuse, alterations, error or failure to properly maintain, clean or repair products.
Researchers at the US Department of Energy's Brookhaven National Laboratory have 
developed a device, dubbed 'ThraxVac,' that can collect and kill anthrax and other 
bacterial spores. The patent-pending device has been licensed to Circle Group Holdings, 
Inc, a public company based in Mundeen, Illinois.
ThraxVac vacuums up anthrax and other bacterial spores, then 'tricks' the spores into 
germinating through heat and moisture, making them vulnerable to injury. The newly 
activated spores are then bombarded with alpha particles, which kill the spores, 
rendering them non-toxic.
Read more: http://www.theengineer.co.uk/news/thraxvac-kills-anthrax-spores/
279013.article#ixzz2cdvYXqRK
MARCAPASSOS COM PU-238
Baterias de Pu-238 + termopar 
Meia-vida 87,7 anos 
Introduzidos em 1973 
Em 1980’s substituídos por baterias a lítio
APLICAÇÕES DE EMISSORES ALFA
! Radioisotope thermoelectric generator (RTG) 
! Pu-238 
! SNAP 3 em 1961 - Navy Transit 4A spacecraft
EMISSÃO BETA 
NEGATIVA
EMISSÃO BETA
! Partícula beta 
! M(beta) = 0,00055 u.m.a. (=igual ao elétron) 
! Carga = -1,6 . 10-19 C 
! Espectro contínuo de energias N=17Z=15
... ...
N=16Z=16
EMISSÃO BETA
! Partícula beta 
! M(beta) = 0,00055 u.m.a. (=igual ao elétron) 
! Carga = -1,6 . 10-19 C 
! Espectro contínuo de energias N=17Z=15
... ...
N=16Z=16
Nêutron " próton + beta neg. + antineutrino
P3215 S3216 + beta + antineutrino
n10 ! p11 + e0�1 + ⌫00
1
ENERGIA DA PARTÍCULA BETA
Kneutrino
Kbeta
Espectro contínuo: 
Energia cinética divide-se 
entre o neutrino e a beta
APLICAÇÕES DA BETA NEGATIVA: 
BETATERAPIA
! Sr-90. Emissor beta puro. 
! Em até  97% dos casos, evita a formação de quelóides.
 THE USE OF STRONTIUM-90 BETA RADIOTHERAPY AS 
ADJUVANT TREATMENT FOR CONJUNCTIVAL MELANOMA
Fonte: http://www.hindawi.com/journals/jo/2013/349162/
http://www.eyephysics.com/ps/ps5/userguide/References/PDF/
Medical_Physics/Astrahan03.pdf