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Q.
The transition in $He ^{+}$ ion that will give rise to a spectral line having the same wavelength as that of some spectral line in hydrogen atom is
Atoms
Solution:
In case of $He ^{+}$ ion the Rydberg constant happens to be four times that in case of hydogen atom. Therefore, to have the same wavelength of some spectral line one must have $\left[\frac{1}{n_{f}^{2}}-\frac{1}{n_{i}^{2}}\right]=4\left[\frac{1}{p_{f}^{2}}-\frac{1}{p_{i}^{2}}\right]$
where $n_{f}$ and $n_{i}$ are principal quantum numbers of final and initial orbits for hydrogen atom and $p_{f}$ and $p_{i}$ are those for $He ^{+}$ ion. This gives $p_{f}=2 n_{f}$ and $p_{i}=2 n_{i}$