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JEE Main 2020 (6 Sep, Shift 2) · Radioactive Decays: Alpha, Beta, Gamma

Radioactive Decays: Alpha, Beta, Gamma question from JEE Main 2020 (6 Sep, Shift 2)

Q21JEE Main 2020 · 6 SEP · SHIFT 2Difficult

Radioactive Decays: Alpha, Beta, Gamma

Given the masses of various atomic particles

mp = 1.0072 u, mn = 1.0087 u, me = 0.000548 u,

mvˉm_{\bar{v}} = 0, md = 2.0141 u, where p ≡\equiv proton, n ≡\equiv neutron,

e ≡\equiv electron, vˉ≡\bar{v} \equiv antineutrino and d ≡\equiv deuteron. Which of the following process is allowed by momentum and energy conservation?

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Answer: (b)

  1. A process is allowed only if the final rest mass is not more than the initial mass and momentum can also be conserved.
  2. (b) mn+mp=2.0159 u>md=2.0141 um_n + m_p = 2.0159\ \text{u} > m_d = 2.0141\ \text{u}: the extra energy goes to the γ\gamma and d recoil, which share momentum. Allowed.
  3. (a) Two particles fusing into one body: in the CM frame the single product must be at rest, so the released energy has nowhere to go. Not allowed.
  4. (c) mn+me>mpm_n + m_e > m_p: needs energy, so a free proton cannot decay. (d) A single photon cannot have zero momentum, so e+e−→γe^+ e^- \to \gamma fails in the CM frame.
  5. So the answer is (b) n+p→d+γn + p \rightarrow d + \gamma.
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