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Atoms and Nuclei question

2021 · 20 Jul · Shift 1 · Q56
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  5. /2021 · 20 Jul · Shift 1 · Q56

Atoms and Nuclei question

2021 · 20 Jul · Shift 1 · Q56

JEE MainPhysicsAtoms and NucleiMCQ+4 / −1
A nucleus of mass M emits γ\gammaγ -ray photon of frequency 'v'. The loss of internal energy by the nucleus is : [Take 'c' as the speed of electromagnetic wave]
  1. A
    hv
  2. B
    hv[1+hv2Mc2]hv\left[ {1 + {{hv} \over {2M{c^2}}}} \right]hv[1+2Mc2hv​]
  3. C
    hv[1−hv2Mc2]hv\left[ {1 - {{hv} \over {2M{c^2}}}} \right]hv[1−2Mc2hv​]
  4. D
    0
View written solutionFree

Correct answer: B

  1. Key idea: recoil of the nucleus

When a nucleus emits a γ\gammaγ-photon, momentum must be conserved. Since the photon carries momentum, the nucleus recoils in the opposite direction.

So, the loss of internal energy of the nucleus is not just hνh\nuhν; it must also provide the recoil kinetic energy of the nucleus.

  1. Energy of emitted photon

The emitted photon has energy

Eγ=hνE_\gamma = h\nuEγ​=hν

and momentum

pγ=hνc.p_\gamma = \frac{h\nu}{c}.pγ​=chν​.
  1. Recoil momentum of nucleus

If the nucleus was initially at rest, then by conservation of momentum,

pnucleus=hνc.p_{\text{nucleus}} = \frac{h\nu}{c}.pnucleus​=chν​.

Hence recoil kinetic energy of the nucleus is

K=p22M=12M(hνc)2=h2ν22Mc2.K = \frac{p^2}{2M} = \frac{1}{2M}\left(\frac{h\nu}{c}\right)^2 = \frac{h^2\nu^2}{2Mc^2}.K=2Mp2​=2M1​(chν​)2=2Mc2h2ν2​.
  1. Loss of internal energy

The total loss of internal energy of the nucleus equals:

  • energy carried by photon, plus
  • recoil kinetic energy of nucleus.

Therefore,

ΔE=hν+h2ν22Mc2.\Delta E = h\nu + \frac{h^2\nu^2}{2Mc^2}.ΔE=hν+2Mc2h2ν2​.

Factor out hνh\nuhν:

ΔE=hν(1+hν2Mc2).\Delta E = h\nu\left(1 + \frac{h\nu}{2Mc^2}\right).ΔE=hν(1+2Mc2hν​).
  1. Compare with options
  • A: hνh\nuhν only ignores recoil, so incorrect.
  • B: hν(1+hν2Mc2)h\nu\left(1+\frac{h\nu}{2Mc^2}\right)hν(1+2Mc2hν​) correct.
  • C: wrong sign.
  • D: impossible.
  1. Final answer

The loss of internal energy by the nucleus is

hν(1+hν2Mc2)\boxed{h\nu\left(1+\frac{h\nu}{2Mc^2}\right)}hν(1+2Mc2hν​)​

which corresponds to Option B.

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