Question

Estimate the thickness of lead (density 11.3 gcm⁻³) required to absorb 90% of gamma rays of energy 1 MeV. The absorption cross-section or gammas of 1 MeV in lead (A = 207) is 20 barns/atom.

Answer 1

Number of gamma rays absorbed in the thickness x cm of lead 

N = N₀(1 − e^{−μ x}) 

where N₀ is the initial number and μ is the absorption coefficient expressed in cm⁻¹. 

Now μ = σ N₀ρ/A where N₀ is the Avagadro’s number, ρ is the density of lead and A is its atomic weight. 

μ = 20 × 10⁻²⁴ × 6.02 × 10²³ × 11.3/207 = 0.657 cm⁻¹ 

n/n₀ = 90/100 = 1 − e^−0.657x 

x = 3.5 cm