A beam of 1.2 MeV photons are incident on a 2 mm lead shield. a) What is the fraction uncollided flux exiting the shield? b) What fraction initially interacts by Photoelectric Effect?

redolrn

redolrn

Answered question

2022-09-05

A beam of 1.2 MeV photons are incident on a 2 mm lead shield.
a) What is the fraction uncollided flux exiting the shield?
b) What fraction initially interacts by Photoelectric Effect?

Answer & Explanation

antidootnw

antidootnw

Beginner2022-09-06Added 10 answers

a) We know that the density of the lead = 11.29  g/cm 3
The mass attenuation coefficient
μ g = 0.125  cm 2 / g
Now, the equation of linear attenuation
I = I 0 e μ x I = I 0 e μ ρ x
and
ϕ = ϕ 0 e μ x ϕ ϕ 0 = e μ ρ x ϕ ϕ 0 = e 0.125 × 11.35 × 2 × 10 3 ϕ ϕ 0 = 0.997
Jannek93

Jannek93

Beginner2022-09-07Added 1 answers

b) The probability of the photoelectric interaction is
p 1 = Z 4 E 35 = Z 4 E 3
The fraction in teracted via photoelectric = Probability × untransmitted particles
= Z 4 E 3 [ 1 0.997 ] = 42386.415 × 10 13 = 4.238 × 10 17

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