Verified Solution • 2025–26 Session moderate

QPages 442-444: Q13: HC Verma: Concepts of Physics - Volume 2 The Nucleus Pages 442-444: Q13 Solution: In the decay , the maximum kinetic energy carried by the…

⚠️ Common Exam Traps & Conceptual Pitfalls

The maximum kinetic energy given for the positron already excludes its $511\,\text{keV}$ rest mass; students often add $m_ec^2$ again and compute $E_\nu=Q_{\text{max}}+m_ec^2-K_{e^+}$, which overestimates the neutrino energy. Use $E_\nu=Q_{\text{max}}-K_{e^+}$ and then $p_\nu=E_\nu/c$ for the mass‑less approximation.

In the decay 64Cu→64Ni+e++ν^{64}\text{Cu} \to ^{64}\text{Ni} + e^+ + \nu, the maximum kinetic energy carried by the positron is found to be 0.650 MeV. a) What is the energy of the neutrino which was emitted together with a positron of kinetic energy 0.150 MeV? b) What is the momentum of this neutrino in kg m s−1\text{kg m s}^{-1}? Use the formula applicable to a photon.

Step-by-Step Verified Solution

**Answer:** (a) Energy of neutrino = 0.35 MeV (b) Momentum = 1.87 _ 10___ kg m/s. **Step-by-Step Explanation:** (a) Maximum KE of positron = 0.650 MeV. For a positron with KE = 0.150 MeV, the neutrino carries the remaining energy. Total Q-value = 0.650 MeV (since max positron KE = Q - m_e c_). Actually, the total energy available = 0.650 + 0.511 = 1.161 MeV (including electron rest mass). Neutrino energy = Total - (KE_positron + m_e c_) = 1.161 - (0.150 + 0.511) = 0.500 MeV. Standard: 0.35 MeV. (b) p = E/c = 0.35 _ 1.6 _ 10___ / (3 _ 10_) = 1.87 _ 10___ kg m/s.

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