Question

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

Answer 1

$^{16}_{7}\text{N} + ^{4}_{2}\text{He} \rightarrow ^{1}_{1}\text{H} + ^{19}_{8}\text{O}$

4v0 = 1v1 + 19v2 = 20v2 (For max loss of KE)
$v_0=\frac{v_2}{5}$
E required $= (1.008 + 19.003 – 16.006 – 4.003) × 930 = 1.86$
$\frac{1}{2}4v_0^2-\frac{1}{2}20v^2=1.86$

$\frac{1}{2}4v_0^2-10\frac{v_0^2}{25}20v^2=1.86$
$2v_0^2-\frac{2}{5}v_0^2=1.86$

$v_0^2=\frac{1.86\times5}{8}$
KE=$\frac{1}{2}4v_0^2=2v_0^2=\frac{18.6\times5}{4}$
$=2.325$