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Concepts Of Physics MCQ Edition [Volume 2]
Exam
Concepts Of Physics MCQ Edition [Volume 2]
Subject
Physics
Chapter
The Nucleus

Concepts Of Physics MCQ Edition [Volume 2] · Physics · The Nucleus

In the decay ^ 64 Cu → ^ 64 Ni + e^ + + , the maximum kinetic...

In the decay ^{64}\text{Cu} \to ^{64}\text{Ni} + e^{+} + \nu,themaximumkineticenergycarriedbythepositronisobservedtobe, the maximum kinetic energy carried by the positron is observed to be0.650 \text{ MeV}.Determinethemomentumoftheneutrinoin. Determine the momentum of the neutrino in\text{kg m s}^{-1}ifitisemittedalongsideapositronhavingakineticenergyofif it is emitted alongside a positron having a kinetic energy of0.150 \text{ MeV}. Use the relation applicable to a photon.

Options

  1. A1.33×10−22 kg m s−11.33 \times 10^{-22} \text{ kg m s}^{-1}
  2. B3.47×10−22 kg m s−13.47 \times 10^{-22} \text{ kg m s}^{-1}
  3. C8.00×10−23 kg m s−18.00 \times 10^{-23} \text{ kg m s}^{-1}
  4. D2.67×10−22 kg m s−12.67 \times 10^{-22} \text{ kg m s}^{-1}

Correct answer

D. 2.67×10−22 kg m s−12.67 \times 10^{-22} \text{ kg m s}^{-1}

Step-by-step solution

The total disintegration energy (QQ-value) is shared between the positron and the neutrino, neglecting the recoil of the daughter nucleus.

Q = K_\max(e^{+}) = 0.650 \text{ MeV}

When the positron has a kinetic energy Ke+=0.150 MeVK_{e^{+}} = 0.150 \text{ MeV}, the energy of the neutrino is:

Eν=Q−Ke+=0.650 MeV−0.150 MeV=0.500 MeVE_{\nu} = Q - K_{e^{+}} = 0.650 \text{ MeV} - 0.150 \text{ MeV} = 0.500 \text{ MeV}

Converting this energy to Joules:

Eν=0.500×106×1.6×10−19 J=8.0×10−14 JE_{\nu} = 0.500 \times 10^{6} \times 1.6 \times 10^{-19} \text{ J} = 8.0 \times 10^{-14} \text{ J}

For a neutrino, the relationship between energy and momentum is the same as that for a photon, Eν=pνcE_{\nu} = p_{\nu} c.

pν=Eνc=8.0×10−143×108 kg m s−1p_{\nu} = \dfrac{E_{\nu}}{c} = \dfrac{8.0 \times 10^{-14}}{3 \times 10^{8}} \text{ kg m s}^{-1}

pν≈2.67×10−22 kg m s−1p_{\nu} \approx 2.67 \times 10^{-22} \text{ kg m s}^{-1}
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