Physics - Modern Physics Question with Solution | TestHub

Let the reaction be represented as ${}_{\mathrm{Z}}^{220}\mathrm{X}⟶\underset{\mathrm{Z}-2}{216}\begin{array}{c}4\\ \mathrm{Y}+\mathrm{H}\text{ e }\\ 2\end{array}$ $\because$ energy released in the reaction is $5\mathrm{MeV}$ $\Rightarrow \frac{1}{2}{m}_Y{v}_Y^2+\frac{1}{2}{m}_{\alpha }{v}_{\alpha }^2=5\mathrm{MeV}$
Also using conservation of linear momentum. $\begin{array}{rl}& m_Y{v}_Y=-m_{\alpha }{v}_{\alpha }\Rightarrow v_Y=\frac{-m_{\alpha }}{m_Y}{v}_{\alpha }\\ & v_Y=\frac{-4}{216}{v}_{\alpha }=\frac{-1}{54}{v}_{\alpha }\end{array}$ Putting in eqn. (i) $\begin{array}{rl}& \frac{1}{2}(216){\left(\frac{v_{\alpha }}{54}\right)}^2+\frac{1}{2}4{\left(v_{\alpha }\right)}^2=5\mathrm{MeV}\\ & \frac{1}{2}(216){\left(\frac{v_{\alpha }}{54}\right)}^2+{\mathrm{K}}_{\mathrm{E}}{\mathrm{E}}_{\alpha }=5\mathrm{MeV}\\ & \Rightarrow \frac{1}{2}\times 4\frac{v_{\alpha }^2}{54}+\mathrm{K}\cdot {\mathrm{E}}_{\cdot \alpha }=5\mathrm{MeV}\\ & \Rightarrow \frac{1}{54}\left({\mathrm{K}}_{\cdot }{\mathrm{E}}_{\alpha }\right)+{\mathrm{K}}_{\cdot }{\mathrm{E}}_{\cdot \alpha }=5\mathrm{MeV}\\ & \text{ K.E. }\alpha =\frac{5\times 54}{55}=\frac{54}{11}\mathrm{MeV}\end{array}$