Question

The minimum kinetic energy needed by an alpha particle to cause the nuclear reaction ${ }_7^{16} N +{ }_2^4 He \rightarrow{ }_1^1 H +{ }_8^{19} O$ in a laboratory frame is $n$ (in $MeV$ ). Assume that ${ }_7^{16} N$ is at rest in the laboratory frame. The masses of ${ }_7^{16} N ,{ }_2^4 He ,{ }_1^1 H$ and ${ }_8^{19} O$ can be taken to be $16.006 \,u, 4.003 \,u$, $1.008 \,u$ and $19.003\, u$, respectively, where $1 u=930\, MeVc ^{-2}$. The value of $n$ is ___

Answer 1

${ }_7^{16} N +{ }_2^4 He \rightarrow{ }_1^1 He +{ }_8^{19} O $
$ { }_7^{16} N +{ }_2^4 He \rightarrow{ }_1^1 He +{ }_8^{19} O $
$ 16.006 \,\,\,\, 4.003 \,\,\,\, 1.008 \,\,\,\, 19.003 $
$ \left.4 v _0=1 v _1+19 v _2=20 v _2 \,\,\,\,\text { (For max loss of } KE \right) $
$ v _0=\frac{ v _2}{5} $
$ E \text { required }=(1.008+19.003-16.006-4.003) \times 930=1.86 $
$ \frac{1}{2} 4 v _0^2-\frac{1}{2} 20 v ^2=1.86$
$ \frac{1}{2} 4 v _0^2-10 \frac{ v _0^2}{25} 20 v ^2=1.86 $
$ 2 v _0^2-\frac{2}{5} v _0^2=1.86 $
$ \frac{8}{5} v _0^2=1.86 $
$ v _0^2=\frac{1.86 \times 5}{8} $
$ KE =\frac{1}{2} 4 v _0^2=2 v _0^2=\frac{18.6 \times 5}{4}$
$ =2.325$