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218 7 QUANTUM THEORY
(i) P = 0.10 W
N = (0.10 W) × (1 s)(700 × 10−9 m)
(6.6261 × 10−34 J s) × (2.9979 × 108ms−1)
= 3.52 × 1017
(ii) P = 1.0 W
N = (1.0 W) × (1 s)(700 × 10−9 m)
(6.6261 × 10−34 J s) × (2.9979 × 108ms−1)
= 3.52 × 1018
E7A.8(b) As described in Section 7A.2 on page 242, photoejection can only occur if the
energy of the incident photon is greater than or equal to the work function of
the metal ϕ. If this condition is ful�lled, the energy of the emitted photon is
given by [7A.10–243], Ek = hν−Φ = hc/λ−Φ. To convert the work function to
Joules, multiply through by the elementary charge, as described in Section 7A.2
on page 242,
Φ = 2.09 eV × e = 2.09 eV × 1.602 × 10−19 J eV−1 = 3.35... × 10−19 J
and since Ek = 1/2meυ2, υ =
√
2Ek/me
(i) For λ = 650 nm
Ephoton =
hc
λ
= (6.6261 × 10−34 J s) × (2.9979 × 108ms−1)
650 × 10−9 m
= 3.06... × 10−19 J
�is is less than the threshold energy, hence no electron ejection occurs.
(ii) For λ = 195 nm
Ephoton =
(6.6261 × 10−34 J s) × (2.9979 × 108ms−1)
195 × 10−9 m
= 1.02... × 10−18 J
�is is greater than the the threshold frequency, and so photoejection can
occur, leading to a kinetic energy of
Ek = hc/λ −Φ = 1.02... × 10−18 J − 3.35... × 10−19 J = 6.84 × 10−19 J
υ =
√
2 × (6.84 × 10−19 J)/(9.109 × 10−31 kg) = 1.23 Mms−1
E7A.9(b) If the power, P, is constant, the total energy emitted in time ∆t is P∆t. �e
energy of each emitted photon is Ephoton = hν = hc/λ. �e total number of
photons emitted in this time period is therefore the total energy emitted divided
by the energy per photon
N = P∆t/Ephoton = P∆tλ/hc