Atoms & Nuclei

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77 of 239

Question 77 of 239

In a hydrogen like atom, when an electron jumps from the M-shell to the L-shell, the wavelength of emitted radiation is LL. If an electron jumps from N\mathrm{N} -shell to the L\mathrm{L} -shell, the wavelength of emitted radiation will be:

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Answer: (D) 2027λ\frac{20}{27} \lambda

When electron jumps from M→L\mathrm{M} \rightarrow \mathrm{L} shell

1λ=K(122−132)=K×536\frac{1}{\lambda}=\mathrm{K}\left(\frac{1}{2^{2}}-\frac{1}{3^{2}}\right)=\frac{\mathrm{K} \times 5}{36}

When eletron jumps from N→L\mathrm{N} \rightarrow \mathrm{L} shell

1λ′=K(122−142)=K×316\frac{1}{\lambda^{\prime}}=K\left(\frac{1}{2^{2}}-\frac{1}{4^{2}}\right)=\frac{K \times 3}{16}

solving equation (i) and (ii) we get

λ′=2027λ\lambda^{\prime}=\frac{20}{27} \lambda

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