A beam of gamma rays is incident on a block of lead. The intensity of gamma rays passing through a uniform medium varies with depth x, in a manner given by the expression I(x) = 1, e-ux Here, I, is the intensity of the radiation at the surface of the material (at x = 0) and u is called the linear absorption coefficient. A) Find an expression for the thickness of material that would absorb half of the gamma rays. B) Find an expression for the thickness of material that would reduce the radiation intensity to a fraction of the initial intensity.

A. x = -㏑(1/2)/μ B. x = -㏑(I/I₀)/μ

Explanation:

A. Since the intensity I = I₀exp(-μx) where I₀ = intensity at x = 0.

When I = I₀/2,  

I = I₀exp(-μx)

I₀/2 = I₀exp(-μx)

dividing through by I₀, we have

1/2 = exp(-μx)

taking natural logarithm of both sides, we have

㏑(1/2) = ㏑[exp(-μx)]

㏑(1/2) = -μx

dividing both sides by -μ

x = -㏑(1/2)/μ

where x is the thickness of the material that would absorb half of the gamma rays

B. Since the intensity I = I₀exp(-μx) at thickness x, where I₀ = intensity at x = 0

I = I₀exp(-μx)

dividing through by I₀

I/I₀ = exp(-μx)

taking natural logarithm of both sides, we have

㏑(I/I₀) = ㏑exp(-μx)

㏑(I/I₀) = -μx

dividing both sides by -μ

x = -㏑(I/I₀)/μ

where x is the thickness of the material that would reduce the radiation intensity to a fraction of the initial intensity.

distance = 150m

displ = root (50 ^2 + 100^2)

root(2500+10000)

root 125000