JEE Advanced 2022 — Paper 1 — PHY-I solution

JEE Advanced2022Paper 1PHY-I

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 _______.

Your answer

Solution

Q-value of the reaction:

$Q = [(m_{N} + m_{H e}) - (m_{H} + m_{O})] c^{2} = (16.006 + 4.003 - 1.008 - 19.003) \times 930$ MeV

$Q = - 1.86$ MeV. Since $Q$ is negative, this much energy must be supplied (endothermic).

Threshold KE in lab frame: The nitrogen target is at rest, so center-of-mass motion carries away kinetic energy that cannot be used for the reaction. The minimum lab-frame KE of the alpha is:

$K E_{m i n} = ∣ Q ∣ \cdot \frac{m _{H e} + m _{N}}{m _{N}} = ∣ Q ∣ \cdot \frac{4 + 16}{16} = \frac{5}{4} \times 1.86 = 2.325$ MeV.

$∴ n \approx 2.33$ .

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