The minimum kinetic energy needed by an alpha particle to cause the nuclear reaction in a laboratory frame is (in MeV). Assume that is at rest in the laboratory frame. The masses of , , and can be taken to be , , and , respectively, where The value

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 $16.006 u$ , $4.003 u$ , $1.008 u$ and $19.003 u$ , respectively, where $1 u = 930 MeV c^{- 2} .$ The value of $n$ is _______. [2022]

Smart Achievers · Accepted Answer

(2.33)

$We have$

$Q = Δ m c^{2}$

$= (m_{N} + m_{He} - m_{H} - m_{O}) \times 930 MeV$

$= (16.006 + 4.003 - 1.008 - 19.003) \times 930 MeV$

$= - 1.86 MeV$ $= 1.86 MeV energy absorbed$

This $Q$ is equal to the maximum loss in kinetic energy of the $α$ -particle.

Considering collision as inelastic, we get maximum loss in

$K . E . = \frac{1}{2} (\frac{4 m \times 16 m}{4 m + 16 m}) v^{2}$

$\Rightarrow Q = (\frac{1}{2} \times 4 m \times v^{2}) \times \frac{16 m}{20 m}$

$\Rightarrow Q = {(K . E .)}_{\min} \times \frac{4}{5}$

$\Rightarrow {(K . E .)}_{\min} = \frac{5}{4} Q = (\frac{5}{4} \times 1.86) MeV = 2.325 MeV$

$∴$ $n = 2.33$

Solution

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