A proton emerges from a slab of water with an energy of 8 Me

A proton emerges from a slab of water with an energy of 8 MeV. It enters the slab with an energy of 25 MeV. How thick is the slab (cm)?

Solution

Hi,

Any gamma or proton when passes through a matter , some part of it gets absorbed, depending on the energy of the incident proton and the thickness of the mater. ( We consider that this attenuation is linear).

Let the incident ray has a energy of I_o and the ray coming out of the slab (matter) be I_x., where x is the thickness of the slab. Now the relationship between these parameters is defined by the formula:

I_x., = I_o e^-x

This formula is called Lambert’s equation and is applicable to linear attenuation. Where = µ x p

µ = mass attenuation coefficient

p= density of the material.

Using this formula let us now solve the given problem. In our problem the I_o = 25MeV

I_x. = 8MeV

We have to find x.

Before this one needs to find the mass attenuation coefficient (µ) of water at 25MeV. This can be obtained from the table found in NIST physics measurement laboratory, maintained by US govt. According to this table ,

The value of mass attenuation coefficient of water at 25mEV is = µ = 1.813 x 10^-2 cm²/g

density of water = 1 g/cm³

SO, = µ x p = 1.813 x 10^-2 cm²/g x 1 g/cm³

         = 1.813 x 10^-2 cm^-1

now, I_x., = I_o e^-x

8MeV = (25MeV)e^{-0.01813 x}

8 /25 = e^{-0.01813 x}

0.32 = e^{-0.01813 x}

x can be solved by taking natural logarithm on both sides,

x = ln(0.32) / -0.01813

x = -1.139 / -0.01813

x= 62.82 cm = thickess of water slab