Question

If the rate of energy loss for a 2.5 MeV proton in graphite of density 2.25 102 kg/m is 32.3 GeV/m, estimate the rate of energy loss in gaseous nitrogen of density 1.29 kg/m?

Answer 1

Answer:185.18 meV/m

Step-by-step explanation:

Energy loss for charge particle is given by

\frac{\mathrm{d} E}{\mathrm{d} x}\propto \frac{\rho Zz^2}{A}

where

\frac{\mathrm{d} E}{\mathrm{d} x} Energy loss per unit length

\rho density

Z atomic number

A atomic mass

z charge of incident particle(for proton it is 1)

For graphite

Z=6

A=12

For gaseous nitrogen

Z=7

A=14

`\frac{\mathrm{d} E}{\mathrm{d} x}=(k\rho Z)/(A)`

For graphite

`32.3=(k* 2.25* 10^2* 6)/(12)--1`

`E'=(k* 1.29* 7)/(14)---2`

Divide 1 and 2

`(32.3)/(E')=(2.25* 10^2* 6* 14)/(12* 7* 1.29)`

E'=185.18 MeV/m