Atoms & Nuclei

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140 of 239

Question 140 of 239

Both the nucleus and the atom of some element are in their respective first excited states. They get de-excited by emitting photons of wavelengths λN,λA{\lambda }_{N},{\lambda }_{A} respectively. The ratio λNλA\dfrac{{\lambda }_{N}}{{\lambda }_{A}} is closest to:

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Answer: (B) 10−6{10}^{-6}

λNλa=EaEN\dfrac{{\lambda }_{N}}{{\lambda }_{a}}=\dfrac{{E}_{a}}{{E}_{N}}, where EaandEN{E}_{a}and{E}_{N} are energies of photons from atom and nucleus respectively. EN{E}_{N} is of the order of MeVMeV and Ea{E}_{a} in few eVeV. So, λNλa=10−6\dfrac{{\lambda }_{N}}{{\lambda }_{a}}={10}^{-6}.

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