Question

The minimum kinetic energy needed by an alpha particle to cause the nuclear reaction ${}_7^{16}N+{}_2^{4}He\to {}_1^{1}H+{}_8^{19}O$ 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 $16.006u,4.003u$, $1.008u$ and $19.003u$, respectively, where $1u=930MeV{c}^{-2}$. The value of $n$ is ___

Answer 1

Answer: 2.33

${}_7^{16}N+{}_2^{4}He\to {}_1^{1}He+{}_8^{19}O$
${}_7^{16}N+{}_2^{4}He\to {}_1^{1}He+{}_8^{19}O$
$16.0064.0031.00819.003$
$4v_0=1v_1+19v_2=20v_2\text{ (For max loss of }KE)$
$v_0=\frac{v_2}{5}$
$E\text{ required }=(1.008+19.003-16.006-4.003)\times 930=1.86$
$\frac{1}{2}4v_0^2-\frac{1}{2}20v^2=1.86$
$\frac{1}{2}4v_0^2-10\frac{v_0^2}{25}20v^2=1.86$
$2v_0^2-\frac{2}{5}{v}_0^2=1.86$
$\frac{8}{5}{v}_0^2=1.86$
$v_0^2=\frac{1.86\times 5}{8}$
$KE=\frac{1}{2}4v_0^2=2v_0^2=\frac{18.6\times 5}{4}$
$=2.325$