resolvendo problemas do livro elements of xray

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Sukhpreet Kaur 
23026113 
MSE 104 
Problem Set 1 
 
1.1 What is the frequency (per second) and energy per quantum(in joules) of x-ray beams of 
wavelength .71 Ǻ (Mo Kα) and 1.5 Ǻ (Cu Kα). 
The frequency of the wave is given by 𝑓 =
𝑐
𝜆
, where f= frequency, c = velocity of light = 
3.00 × 108 𝑚/𝑠𝑒𝑐, and 𝜆 = 𝑤𝑎𝑣𝑒𝑙𝑒𝑛𝑔𝑡ℎ. 
So, 𝑓 =
𝑐
𝜆
 = 
3.00×108 
𝑚
s
.71 ×10−10𝑚
 = 𝟒. 𝟐𝟑 × 𝟏𝟎𝟏𝟖 𝒔−𝟏 
 And 𝑓 =
𝑐
𝜆
 = 
3.00×108 
𝑚
s
1.5 ×10−10𝑚
 = 𝟏. 𝟗𝟓 × 𝟏𝟎𝟏𝟖 𝒔−𝟏 
 
Energy = E = hf where h= 6.63 × 10−34 𝑗𝑜𝑢𝑙𝑒𝑠 ∗ 𝑠 
So, Energy of Mo= 𝐸 = ( 6.63 × 10−34 𝑗𝑜𝑢𝑙𝑒𝑠 ∗ 𝑠 ) × 4.23 × 1018 𝑠−1 = 𝟐. 𝟖 ×
𝟏𝟎−𝟏𝟓𝒋𝒐𝒖𝒍𝒆𝒔 
Energy of Cu= 𝐸 = ( 6.63 × 10−34 𝑗𝑜𝑢𝑙𝑒𝑠 ∗ 𝑠 ) × 1.98 × 1018 𝑠−1 = 𝟏. 𝟐𝟗 × 𝟏𝟎−𝟏𝟓𝒋𝒐𝒖𝒍𝒆𝒔 
 
 
 
 
 
 
 
 
 
 
 
 
1.3 Show that the velocity with which electrons strike the target of an x-ray tube depends only on 
the voltage between anode (target) and cathode and not on the distance between them. [The 
force on a charge e (coulombs) by a field E (volts/m) in eE newtons. ] 
The kinetic energy of the electrons on impact with the target is, 
𝐾𝐸 = 𝑒𝑉 =
1
2
𝑚𝑣2 
 where e= charge on electron, V is the voltage, m is the mass of the electro and v is the 
velocity. 
 In x-ray tube, the high voltage draws the electrons to the target/anode, which they strike 
with very high velocity. The tube is a vacuum tube, thus the electrons do not feel inference of 
any other force accept that from the voltage. Since this is true, electron will stay in constant 
motion and will not accelerate over distance, so variance in distance doesn’t matter. When the 
electrons strike the anode, x rays are produced at the point of impact. 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
1.6 Graphically verify Eq. (1-13) for a lead absorber and Mo Kα, Rh Kα and Ag Kα radiation. ( 
The mass absorption coefficients of lead for these radiations are 122.8, 84.13 and 66.14 cm^2/g, 
respectively). From the curve, determine the mass absorption coefficient of lead for the shortest 
wavelength radiation from a tube operated at 30,000 volts. 
 
Equation 1-13 is: 
 
𝜇
𝜌
= 𝑘 λ3𝑍3 
 
 Where 
𝜇
𝜌
 is the mass absorption coefficient, 𝜌 is the density and Z is the atomic number of the 
absorber. 
The wavelengths, along with the absorption edges ( .14 and .98) of Pb, were found in from 
Appendix 7. By using the absorption edge and the data given, the following graph was produced: 
 
 
The curve produced is roughly similar to how the graph of the equation y^3 would look like. 
Thus it could be seen that the Eq 1-13 is right. 
The shortest wavelength limit can be found by using the following equation: 
𝜆𝑆𝑊𝐿 =
1.24 × 103
𝑉
, 𝑤ℎ𝑒𝑟𝑒 𝑉 = 𝑣𝑜𝑙𝑡𝑎𝑔𝑒 
So, for 30,000 volts, 
 𝜆𝑆𝑊𝐿 =
1.24 ×103
30000
= .4133 Ǻ 
0
20
40
60
80
100
120
140
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8
M
as
s 
ab
so
rp
ti
o
n
 c
o
ef
fi
ci
en
t
Wavelength
Mass Absorption Coefficient and Wavelength plot
 
To get the mass absorption coefficient for this wavelength, interpolate from the above mass. 
 
 
𝜇
𝜌
 ~ 45 cm^2 / g 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
1.8 (a) Calculate the mass and linear absorption coefficients of air for Cr Kα radiation. Assume 
that air contains 80% nitrogen and 20 % oxygen by weight and has a density of 1.29 ×
10−3
𝑔
𝑐𝑚^3
 . (b) Plot the transmission factor of air for Cr Kα radiation and a path length of 0 to 20 
cm. 
(a) Air is composed of more than one element. In this case, the mass absorption coefficient is the 
weighted average of the mass absorption coefficients of the two elements it’s composed of. 
According to 1-12. 
𝜇
𝜌
= 𝑤1 (
𝜇
𝜌
)
1
+ 𝑤2 (
𝜇
𝜌
)
2
+ ⋯ 
Since air contains 80% nitrogen and 20 % oxygen, 
 
𝜇
𝜌𝑎𝑖𝑟
= (. 8) (
𝜇
𝜌
)
𝑁
+ (. 2) (
𝜇
𝜌
)
𝑂
= (. 8)(24.42) + (. 2)(37.19) = 26.97
𝑐𝑚2
𝑔𝑚
 
 The mass absorption values of N and O for Cr Ka radiation were found from the 
Appendix 8 of the textbook. 
The linear absorption coefficient is 
 𝜇 = (26.97
𝑐𝑚2
𝑔
) ( 1.29 ×
10−3𝑔
𝑐𝑚3
) = 3.48 × 10−2𝑐𝑚−1 
 
(b) When x-rays come in contact with matter, they are transmitted and partly absorbed. 
Intensity of the x-ray beam as it passes through an object tends to decrease proportionally to the 
distance traveled x. The transmission factor is the ratio of final intensity , 𝐼𝑥, after penetrating the 
length x over the initial, incident intensity, 𝐼0. 
It’s given by the following equation: 
𝐼𝑥
𝐼0
= 𝑒−𝜇𝑥 
The linear absorption coefficient was calculated earlier, and by using various path lengths 
between 0 to 20 cm, the following graph was generated. The following graph plots the 
transmission factor of air in relation to the path lengths. As can be seen from the graph below, 
the transmission factor decreased as the path length increased. 
 
 
0
0.2
0.4
0.6
0.8
1
1.2
0 5 10 15 20 25
Tr
an
sm
is
si
o
n
 F
ac
to
r
Path Length
Transmission factor of air for Cr Kα radiation and 
vs Path length Plot.