Question:
Hydrogen (\({}_1\mathrm{H}^1\)), Deuterium (\({}_1\mathrm{H}^2\)), singly ionised Helium (\(({}_2\mathrm{He}^4)^+\)) and doubly ionised lithium (\(({}_3\mathrm{Li}^6)^{++}\)) all have one electron around the nucleus. Consider an electron transition from \(n=2\) to \(n=1\). If the wave lengths of emitted radiation are \(\lambda_1,\ \lambda_2,\ \lambda_3\) and \(\lambda_4\) respectively then approximately which one of the following is correct ?
Hydrogen (\({}_1\mathrm{H}^1\)), Deuterium (\({}_1\mathrm{H}^2\)), singly ionised Helium (\(({}_2\mathrm{He}^4)^+\)) and doubly ionised lithium (\(({}_3\mathrm{Li}^6)^{++}\)) all have one electron around the nucleus. Consider an electron transition from \(n=2\) to \(n=1\). If the wave lengths of emitted radiation are \(\lambda_1,\ \lambda_2,\ \lambda_3\) and \(\lambda_4\) respectively then approximately which one of the following is correct ?
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For hydrogen-like ions \(\frac1\lambda=RZ^2\left(\frac1{n_1^2}-\frac1{n_2^2}\right)\), so \(\lambda\propto\frac1{Z^2}\) (nuclear mass effect is negligible).
Updated On: Nov 11, 2026
- \(4\lambda_1=2\lambda_2=2\lambda_3=\lambda_4\)
- \(\lambda_1=2\lambda_2=2\lambda_3=\lambda_4\)
- \(\lambda_1=\lambda_2=4\lambda_3=9\lambda_4\)
- \(\lambda_1=2\lambda_2=3\lambda_3=4\lambda_4\)
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The Correct Option is C
Solution and Explanation
For \(2\to1\): \(\frac1\lambda=RZ^2\left(1-\frac14\right)=\frac34RZ^2\), so \(\lambda\propto\frac1{Z^2}\).
H and D both have \(Z=1\): \(\lambda_1\approx\lambda_2\) (difference only via the tiny reduced-mass correction).
\(\mathrm{He}^+\) has \(Z=2\): \(\lambda_3=\frac{\lambda_1}{4}\). \(\mathrm{Li}^{++}\) has \(Z=3\): \(\lambda_4=\frac{\lambda_1}{9}\).
So \(\lambda_1=\lambda_2=4\lambda_3=9\lambda_4\).
Answer: (3) \(\lambda_1=\lambda_2=4\lambda_3=9\lambda_4\)
H and D both have \(Z=1\): \(\lambda_1\approx\lambda_2\) (difference only via the tiny reduced-mass correction).
\(\mathrm{He}^+\) has \(Z=2\): \(\lambda_3=\frac{\lambda_1}{4}\). \(\mathrm{Li}^{++}\) has \(Z=3\): \(\lambda_4=\frac{\lambda_1}{9}\).
So \(\lambda_1=\lambda_2=4\lambda_3=9\lambda_4\).
Answer: (3) \(\lambda_1=\lambda_2=4\lambda_3=9\lambda_4\)
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