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JEE PHYSICS MCQ: NUCLEI

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Application +4 | -1
In the decay 64Cu→64Ni+e++v{ }^{64} {Cu} \rightarrow{ }^{64} {Ni}+e^{+}+{v}, the maximum kinetic energy carried by the positron is found to be 0.650 MeV0.650 \mathrm{~MeV}.

(a) What is the energy of the neutrino which was emitted together with a positron of kinetic energy 0.150 MeV?0.150 \mathrm{~MeV}?
(b) What is the momentum of this neutrino in kg m s−1?\mathrm{kg} \mathrm{~m} \mathrm{~s}^{-1}?

Use the formula applicable to a photon.
View answer and explanation
Correct option: A
(a) 500 keV;500 \mathrm{~keV}; (b) 2.67×10−22 kg m s−12.67 \times 10^{-22} \mathrm{~kg} \mathrm{~m} \mathrm{~s}^{-1}
Option A ((a) 500 keV;500 \mathrm{~keV}; (b) 2.67×10−22 kg m s−12.67 \times 10^{-22} \mathrm{~kg} \mathrm{~m} \mathrm{~s}^{-1}) is correct because the neutrino energy is the difference between the maximum positron energy and the given positron kinetic energy, and the momentum is calculated using the photon-like relationship.

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