Question

A total of 1012 electrons, each with 100 MeV of kinetic energy, are incident on a 1 cm by 1 cm square piece of graphite (Z = 6, A = 12, density = 1.8 g/cm3). The thickness of the graphite is equal to the range of 100 MeV electrons in graphite.

a) what is the thickness of graphite in cm?

b) if all of the Bremsstrahlung radiation produced by the 100 MeV electrons escapes the graphite, and if all of the delta electrons stop in the graphite, how much energy per gram is deposited by the 10^12 electron stopping in graphite?

c) If the collisional mass stopping power of 100 MeV in graphite is 1.95 MeV∙cm2/g, what is the total mass stopping power?

b) what thickness of graphite is needed to reduce the electron energy from 100 MeV down to 80 MeV?

Answer 1

The Kinetic energy of each electron is 100 MeV.

The information provided is,

Z = 6, A = 12 and density = 1.8g/cm3.

a) The thickness of graphite is equal to the range of 100 MeV electrons in graphite.

Volume = 12 / 1.8 = 6.67 cm3

Surface Area = 1 cm2

Thickness = 6.67 cm.

b) As per the calculation in (c),

Total mass stopping power of an electron is 97.5 MeV cm2/g

for 10^12 electrons, E = 97.5 MeV cm2/g * 10^12 * 2.99 * 10^10 is the energy

Therefore, E = 291.525 * 10^22.

c) As we know the collisional mass stopping power is approximately 2% of the total mass stopping power.

Here collisional mass stopping power is 1.95 MeV cm2/g, which is 2% of the total mass stopping power.

Total mass stopping power = 1.95MeV cm2/g * 100/2 = 195/2 MeV cm2/g = 97.5 MeV cm2/g

d) The Thickness does not change with the energy, Hence the thickness remains even if energy changes from 100 MeV down to 80 MeV