Question

When a neutron collides with a quasi-free proton, it loses half of its energy on the average in every collision. How many collisions, on the average, are required to reduce a $2MeV$ neutron to thermal energy df $0.04eV.$

• A. $30$
• B. $22$
• C. $35$
• D. $26$

Answer 1

The correct option is D $26$

Let $E_0$ be the initial energy of neutron, the energy of neutron after 1 collision reduces to $E_0/2=E_1\left(let\right)$ i.e. $E_1/E_0=1/2$.

After second collision, $E_2/E_0=(1/2{)}^2$, therefore after $n$ collision.

$\frac{E_n}{E_0}=(\frac{1}{2}{)}^n$

Here, given $E_0=2MeV,E_n=0.04eV=0.04\times {10}^{-6}MeV$

Hence, $\frac{0.04\times {10}^{-6}}{2}=(\frac{1}{2}{)}^n$

$2\times {10}^{-8}=(\frac{1}{2}{)}^n$

$log2-8log10=-nlog2$

$0.3010-8=-0.3010n$

$n=0.7699/0.3010=25.58$