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(D) The Rydberg formula for a hydrogen-like species is given by $\frac{1}{\lambda} = R Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$.For $He^{

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$n = 2$ to $n = 1$

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(D) The Rydberg formula for a hydrogen-like species is given by $\frac{1}{\lambda} = R Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$.For $He^{+}$ $(Z = 2)$,the transition from $n_2 = 4$ to $n_1 = 2$ gives $\frac{1}{\lambda} = R (2)^2 \left( \frac{1}{2^2} - \frac{1}{4^2} \right) = 4R \left( \frac{1}{4} - \frac{1}{16} \right) = 4R \left( \frac{3}{16} \right) = \frac{3R}{4}$.For a hydrogen atom $(Z = 1)$,we need to find $n_1$ and $n_2$ such that $\frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) = \frac{3R}{4}$.Comparing the two,we get $\frac{1}{n_1^2} - \frac{1}{n_2^2} = \frac{3}{4}$.This is satisfied when $n_1 = 1$ and $n_2 = 2$ because $\frac{1}{1^2} - \frac{1}{2^2} = 1 - \frac{1}{4} = \frac{3}{4}$.

Which transition in the hydrogen atomic spectrum will have the same wavelength as the transition,$n = 4$ to $n = 2$ of $He^{+}$ spectrum?

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