Calculate the minimum kinetic energy needed by an alpha particle to cause the nuclear reaction ${ }_{\text{7}}^{\text{16}}{\text{N+}}_{\text{2}}^{\text{4}}\text{He}{\to }_{\text{1}}^{\text{1}}{\text{H+}}_{\text{8}}^{\text{19}}\text{O}$  in a laboratory frame (in MeV). Assume that ${ }_{\text{7}}^{\text{16}}\text{N}$   is at rest in the laboratory frame. The masses of  ${ }_{\text{7}}^{\text{16}}{\text{N,}}_{\text{2}}^{\text{4}}{\text{H\hspace{0.17em}and\hspace{0.17em}}}_{\text{8}}^{\text{19}}\text{O}$  can be taken to be  $\text{16.006u,4.003u,1.008u\hspace{0.17em}and\hspace{0.17em}19.003u}$, respectively, where ${\text{1u=930MeVe}}^{\text{-2}}$. 

The value of n is __________.

The given nuclear reaction is
 ${ }_7^{16}N{+}_2^{4}He{\to }_1^{1}H{+}_8^{19}O$
As minimum kinetic energy is needed for incident alpha particle, we consider energy loss in collision to be maximum to provide the energy required for the reaction to occur. For this case we consider  ${ }_7^{19}N$  and  ${ }_2^{4}He$  will move together as the case of inelastic collision. If incident speed of  ${ }_2^{4}He$  is $u$   then after collision with  $N$, final speed $v_1$  will become
 $4m{v}_0+0=(4+16)m{v}_1$
 $v_1=\frac{u}{5}$
Energy required to initiate the reaction is given as
 $E=(1.008+19.003-16.006-4.003)\times 930=1.86MeV$
This energy is provided by the loss of kinetic energy in collision which is given as
$\frac{1}{2}\left(4m\right)u^2-\frac{1}{2}\left(20m\right){\left(\frac{u}{5}\right)}^2=1.86$ 
 $\frac{8}{5}m{u}^2=1.86$
 $m{u}^2=\frac{1.86\times 5}{8}$
Kinetic energy of alpha particle is
 $K_{\alpha }=\frac{1}{2}\left(4m\right)u^2=\frac{18.6\times 5}{4}$
 $K_{\alpha }=2.325MeV$