Solution:

Let initial speed of neutron is $$v_0$$ and kinetic energy is K.

$$1^{st}$$ collision :

by momentum conservation <\br> $$mv_0 = mv_1 + 2 mv_2 \, \Rightarrow \, v_1 + 2v_2 = v_0$$ by $$e = 1$$ $$v_2 - v_1 = v_0$$

$$\Rightarrow \, \, v_2 = \frac{2v_0}{3} ; v_1 = - \frac{v_0}{3} $$

fractional loss $$\frac{\frac{1}{2}mv_{0}^{2} - \frac{1}{2} m\left(\frac{v_{0}}{3}\right)^{2}}{\frac{1}{2}mv^{2}_{0}} $$

$$\Rightarrow P_{b} = \frac{8}{9} \approx.89 $$

$$2^{nd}$$ collision :

by momentum conservation

$$mv_0 = mv_1 + 12 mv_2$$

$$\Rightarrow \, v_1 + 12 v_2 = v_0$$

by $$e = 1 \, \, \, v_2 - v_1 = v_0$$

$$v_2 = \frac{2v_0}{13} ; v_1 = \frac{ -11 v_0}{13}$$

Now fraction loss of energy

$$\frac{P_{c} = \frac{1}{2} mv_{0}^{2} - \frac{1}{2} m \left(\frac{11v_{0}}{13}\right)^{2}}{\frac{1}{2} mv_{0}^{2}} = \frac{48}{169} \approx 0.28 $$