Physics · Nuclear Physics

JEE Advanced 2022 — Paper 1 — Question 20

The minimum kinetic energy needed by an alpha particle to cause the nuclear reaction

${ }_{7}^{16} N+{ }_{2}^{4} \mathrm{He} \rightarrow{ }_{1}^{1} \mathrm{H}+{ }_{8}^{19} \mathrm{O}$ in a laboratory frame is n (in MeV ).

Assume that ${ }_{7}^{16} N$ 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 $16.006 \mathrm{u}, 4.003 \mathrm{u}, 1.008$

$u$ and 19.003 u , respectively, where $1 \mathrm{u}=930 \mathrm{MeVc}^{-2}$ . The value of n is $\qquad$ .

Answer: 2.32

Numerical answer — enter this value.

${ }_{7}^{14} \mathrm{~N}+{ }_{2}^{4} \mathrm{He} \rightarrow{ }_{1}^{1} \mathrm{H}+{ }_{8}^{19} \mathrm{O}$

$\mathrm{Q}=[16.006+4.003-1.008-19.003] \times 930$

$=-1.86 \mathrm{MeV}$

$\mathrm{E}_{\mathrm{th}}=\left(1+\frac{m}{M}\right)|Q| \approx\left(1+\frac{4}{16}\right)(1.86)=2.32 \mathrm{MeV}$

Answer key and solution verified before publishing.

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