QJEEAdv22Phd7ae: JEE Advanced 2022 Physics - Class 12 JEEAdv22Phd7ae Solution: The minimum kinetic energy needed by an alpha particle to cause …

Question

Verified Solution • 2025–26 Session moderate

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

M e V

. 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 \,\mathrm{MeVc}^{-2}

. The value of

n

is ________ .

Accepted Answer

\begin{aligned} Q & = (m_{N} + m_{He} - m_{H} - m_{O}) \times c^{2} \\ = (16.006 + 4.003 - 1.008 - 19.003) \times 930 MeV \\ = - 1.86 MeV \\ = 1.86 MeV energy absorbed. \end{aligned}

And,

\frac{1}{2} \times \frac{m \times 4 m}{5 m} \times v^{2}=

max loss in kinetic energy

\begin{aligned} \Rightarrow \frac{1}{2} m v^{2} & = \frac{5}{4} \times Q \\ = \frac{5}{4} \times (1.86) MeV \\ = 2.325 MeV \end{aligned}

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