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Atoms and Nuclei question

2012 · Shift 0 · Q47
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Atoms and Nuclei question

2012 · Shift 0 · Q47

JEE MainPhysicsAtoms and NucleiMCQ+4 / −1
Assume that a neutron breaks into a proton and an electron. The energy released during this process is : (mass of neutron =1.6725×10−27kg,= 1.6725 \times {10^{ - 27}}kg,=1.6725×10−27kg, mass of proton =1.6725×10−27 kg,= 1.6725 \times {10^{ - 27}}\,kg,=1.6725×10−27kg, mass of electron =9×10−31 kg= 9 \times {10^{ - 31}}\,kg=9×10−31kg ).
  1. A
    0.51MeV0.51MeV0.51MeV
  2. B
    7.10 MeV7.10\,MeV7.10MeV
  3. C
    6.30 MeV6.30\,MeV6.30MeV
  4. D
    5.4 MeV5.4\,MeV5.4MeV
View written solutionFree

Correct answer: A

  1. Use mass-energy equivalence

If a neutron breaks into a proton and an electron, then the energy released is due to the mass defect:

Δm=mn−(mp+me)\Delta m = m_n - (m_p + m_e)Δm=mn​−(mp​+me​)

and

E=Δm c2E = \Delta m \, c^2E=Δmc2
  1. Substitute the given masses

Given:

mn=1.6725×10−27 kgm_n = 1.6725 \times 10^{-27}\,\text{kg}mn​=1.6725×10−27kg mp=1.6725×10−27 kgm_p = 1.6725 \times 10^{-27}\,\text{kg}mp​=1.6725×10−27kg me=9×10−31 kgm_e = 9 \times 10^{-31}\,\text{kg}me​=9×10−31kg

Now,

Δm=1.6725×10−27−(1.6725×10−27+9×10−31)\Delta m = 1.6725 \times 10^{-27} - \left(1.6725 \times 10^{-27} + 9 \times 10^{-31}\right)Δm=1.6725×10−27−(1.6725×10−27+9×10−31) Δm=−9×10−31 kg\Delta m = -9 \times 10^{-31}\,\text{kg}Δm=−9×10−31kg

This gives a negative value, which would mean energy is absorbed, not released. So the given neutron and proton masses are evidently intended to be approximately equal, and the usual neutron-proton mass difference should be used as implied by the option set.

  1. Physical interpretation

For neutron decay,

n→p+e−+νˉn \to p + e^- + \bar\nun→p+e−+νˉ

The actual mass difference between neutron and proton is about the electron rest mass plus a small extra amount. The released kinetic energy is about:

Q≈0.78 MeVQ \approx 0.78\,\text{MeV}Q≈0.78MeV

If one ignores the neutrino and focuses on the electron rest-energy scale from the given approximate data, the closest expected exam answer is the electron rest energy:

Ee=mec2E_e = m_e c^2Ee​=me​c2 =9×10−31×(3×108)2= 9 \times 10^{-31} \times (3 \times 10^8)^2=9×10−31×(3×108)2 =9×10−31×9×1016= 9 \times 10^{-31} \times 9 \times 10^{16}=9×10−31×9×1016 =8.1×10−14 J= 8.1 \times 10^{-14}\,\text{J}=8.1×10−14J

Convert to eV:

1 eV=1.6×10−19 J1\,\text{eV} = 1.6 \times 10^{-19}\,\text{J}1eV=1.6×10−19J

So,

E=8.1×10−141.6×10−19≈5.06×105 eVE = \frac{8.1 \times 10^{-14}}{1.6 \times 10^{-19}} \approx 5.06 \times 10^5\,\text{eV}E=1.6×10−198.1×10−14​≈5.06×105eV E≈0.506 MeV≈0.51 MeVE \approx 0.506\,\text{MeV} \approx 0.51\,\text{MeV}E≈0.506MeV≈0.51MeV
  1. Check options
  • A: 0.51 MeV0.51\,\text{MeV}0.51MeV ✔
  • B: 7.10 MeV7.10\,\text{MeV}7.10MeV ✘
  • C: 6.30 MeV6.30\,\text{MeV}6.30MeV ✘
  • D: 5.4 MeV5.4\,\text{MeV}5.4MeV ✘

Hence, the intended answer is:

0.51 MeV\boxed{0.51\,\text{MeV}}0.51MeV​
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