calculate the mass collision stopping power for a 20-MeV proton in lead, without the shell correction....

calculate the mass collision stopping power for a 20-MeV proton in lead, without the shell correction.

Answer: 11.6 MeV cm^2/g

Solutions

Expert Solution

Bethe's formula for stopping power of charged particle

z =1 for proton , charge on the charged particle

I_ev - average excitation energy of the medium

= 52.8ev + 8.71*z ev , approx.

= 52.8ev + 8.71*82 = 767 ev ( z= 82 for lead)

= v/c , v is the speed of the particle

relativistic kinetic energy of proton mc² / (1-² )^1/2 - mc² = 20 Mev

mc² = 931.5 Mev , rest mass of proton

² = 0.0416

F() = 10.65

n: electron density of lead

density of Pb = 11.34 gm/cc

z= 82 ; 82 electrons per atom 1 mol of Pb = 207 gm

per gm = 6.02e+23/207 * 82 = 2.385e+23 electrons /gm

per cu.m n= 2.385e+23 *11.34e+6 = 2.704e+30

dE/dx = 5.08e-31 *1*2.704e+30 / 0.0416 * (10.65 - ln(767) ) = -131.55 Mev/cm

- linear stopping power

mass stopping power = 1/ * dE/dx = 131.55/11.34 = 11.60 Mev-cm²/gm

genius_generous answered 1 year ago

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