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

JEE Advanced · 2022 · 28 Aug 2022 · Paper 1 · Modern Physics

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

16.006 u, 4.003 u

1.008 u

and

19.003 u

, respectively, where

1 u = 930 M e V c^{- 2}

. The value of

n

is ____.

Answer: 2.325

Q-value of the reaction:

Q = (16.006 + 4.003 - 1.008 - 19.003) \times 930 = 1.86

MeV (endothermic).

Using conservation of momentum and energy in the center-of-mass frame, the threshold kinetic energy is:

K E_{threshold} = \frac{1}{2} \times 4 \times v_{0}^{2}

From energy conservation:

\frac{1}{2} (4) v_{0}^{2} - \frac{1}{2} (20) (\frac{v _{0}}{5})^{2} = 1.86

Simplifying:

2 v_{0}^{2} - \frac{2 v _{0}^{2}}{5} = 1.86

, giving

\frac{8 v _{0}^{2}}{5} = 1.86

Therefore:

K E = 2 v_{0}^{2} = 2 \times \frac{1.86 \times 5}{8} = 2.325

MeV

JEE Advanced · 2022 · 28 Aug 2022 · Paper 1 · Numerical answer