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Dual Nature of Radiation question

2018 · 16 Apr · Shift 1 · Q49
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Dual Nature of Radiation question

2018 · 16 Apr · Shift 1 · Q49

JEE MainPhysicsDual Nature of RadiationMCQ+4 / −1
Both the nucleus and the atom of some element arein their respective first excited states. They get de-excted by emitting photons of wavelengths λ\lambdaλ N, λ\lambdaλ A respectively. The ratio λNλA{{{}^\lambda N} \over {{}^\lambda A}}λAλN​ is closest to :
  1. A
    10 −-− 6
  2. B
    10
  3. C
    10 −-− 10
  4. D
    10 −-− 1
View written solutionFree

Correct answer: A

  1. Use photon energy relation

For de-excitation by emission of a photon, E=hcλE = \frac{hc}{\lambda}E=λhc​ So wavelength is inversely proportional to the transition energy: λ∝1E\lambda \propto \frac{1}{E}λ∝E1​

Thus, λNλA=EAEN\frac{\lambda_N}{\lambda_A} = \frac{E_A}{E_N}λA​λN​​=EN​EA​​ where:

  • ENE_NEN​ = nuclear excitation energy
  • EAE_AEA​ = atomic excitation energy
  1. Estimate typical excitation energies

  • For an atom, the first excited state is typically of order a few eV, say EA∼1 to 10 eVE_A \sim 1\text{ to }10\ \text{eV}EA​∼1 to 10 eV

  • For a nucleus, the first excited state is typically of order keV to MeV, commonly around EN∼105 to 106 eVE_N \sim 10^5\text{ to }10^6\ \text{eV}EN​∼105 to 106 eV

A reasonable order-of-magnitude comparison is: EAEN∼1 eV106 eV=10−6\frac{E_A}{E_N} \sim \frac{1\ \text{eV}}{10^6\ \text{eV}} = 10^{-6}EN​EA​​∼106 eV1 eV​=10−6

Hence, λNλA∼10−6\frac{\lambda_N}{\lambda_A} \sim 10^{-6}λA​λN​​∼10−6

  1. Match with options

The closest option is: 10−610^{-6}10−6

So the correct option is A.

  1. Answer check with stored answer

Stored correct answer: A

This matches our derived answer.

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