It is found that if a neutron suffers an elastic collinear collision with a deuterium at rest, the fractional loss of its energy is P d , while for its similar collision with a carbon nucleus at rest, the fractional loss of energy is P c . The values of P d and P c are respectively

Question

Aicharya · Accepted Answer

Since, the linear momentum is conserved,

u = V_{1} + 2 V_{2}

.

e = 1 = \frac{V_{2} - V_{1}}{u}

(where

e \Rightarrow

the coefficient of restitution),

u = V_{2} - V_{1}

V_{2} = \frac{2 u}{3}; V_{1} = - \frac{u}{3}

. Initial energy,

= \frac{1}{2} u^{2}

. Final energy,

= \frac{1}{2} \times 1 \times \frac{u^{2}}{9} = \frac{u^{2}}{18}

. Fractional change

= \frac{\frac{u^{2}}{2} - \frac{u^{2}}{18}}{\frac{u^{2}}{2}}

= 0.88

u = V_{1} + 12 V_{2}

\Rightarrow 1 = \frac{V_{2} - V_{1}}{u}

\Rightarrow u = V_{2} - V_{1}

\Rightarrow V_{2} = \frac{2 u}{13}

\Rightarrow V_{1} = - \frac{11 u}{13}

.
Change in energy,

= u^{2} - {(\frac{11 u}{13})}^{2} \times 100

= 0.294

It is found that if a neutron suffers an elastic collinear collision with a deuterium at rest, the…

Solution