Question

On an average, a neutron loses half of its energy per collision with a quasi-free proton. To reduce a 2 MeV neutron to a thermal neutron having energy 0.04 eV, the number of collisions required is nearly:

• A. 50
• B. 52
• C. 26
• D. 15

Answer 1

The correct option is C 26

Initial energy$=E_o=2MeV=2\times {10}^{6}eV$

After first collision, $E_1=\frac{E_o}{2}$

After second collision, $E_2=\frac{E_1}{2}=\frac{E_o}{2^2}$

After third collision, $E_3=\frac{E_2}{2}=\frac{E_o}{2^3}$

And so ....

After n collisions- $E_n=\frac{E_o}{2^n}$

And final energy$=0.04eV$

Hence, $E_n=0.04$

$⟹\frac{2\times {10}^6}{2^n}=0.04$

$⟹2^n=\frac{2\times {10}^6}{0.04}=5\times {10}^7$

$⟹n=lo{g}_2(5\times {10}^7)=\frac{lo{g}_{10}(5\times {10}^7)}{{\log }_{10}2}$

$⟹n=\frac{7+0.699}{0.301}$

$⟹n=26$

Answer-(C)