A nucleus with mass number 220 initially at rest emits an alpha particle. If the \(Q\) value of reaction is \(5.5 \mathrm{MeV}\). Calculate the value of kinetic energy of alpha particle

Given, \(Q=5.5 \mathrm{MeV}\)
Mass number of nucleus \(=220\)
By conservation of linear momentum.
\(M_\alpha v_\alpha=M_y v_y\)
where, \(M_\alpha, v_\alpha\) are mass and velocity of \(\alpha\)-particle, and \(M_X v_X\) are mass and velocity of the daughter nuclei \(X\).
Since, \(Q=(\mathrm{KE})_\alpha+(\mathrm{KE})_X\)
Let \(Y\) be the nucleus with the mass number 220 and \(X\) be the daughter nuclei produced.
Hence, we have \({ }^{220} Y \longrightarrow{ }^{216} X+{ }_4^2 \alpha\)
Using conservation of linear momentum.
\(M_\alpha v_\alpha=M_X v_X\)
\(\Rightarrow \quad v_x=\frac{M_\alpha v_\alpha}{M_X}\)
\(\begin{aligned} & \therefore \text { From Eq. (i), } Q=\frac{1}{2} M_\alpha v_\alpha^2+\frac{1}{2} M_X v_X^2 \\ & \Rightarrow \quad 55=\frac{1}{2} M_\alpha v_\alpha^2+\frac{1}{2} \frac{M_\alpha^2 v_\alpha^2}{M_x} \\ & \Rightarrow \quad 55=\frac{1}{2} M_\alpha v_\alpha^2\left[1+\frac{M_\alpha}{M_X}\right] \\ & \Rightarrow \quad 55=(\mathrm{KE})_\alpha\left(1+\frac{4}{216}\right) \\ & \Rightarrow \quad \mathrm{KE}_\alpha=54 \mathrm{MeV}\end{aligned}\)