Answer:185.18 meV/m
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
[tex]\frac{\mathrm{d} E}{\mathrm{d} x}=\frac{k\rho Z}{A}[/tex]
For graphite
[tex]32.3=\frac{k\times 2.25\times 10^2\times 6}{12}--1[/tex]
[tex]E'=\frac{k\times 1.29\times 7}{14}---2[/tex]
Divide 1 and 2
[tex]\frac{32.3}{E'}=\frac{2.25\times 10^2\times 6\times 14}{12\times 7\times 1.29}[/tex]
E'=185.18 MeV/m
