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(D) The wave number for the transition of $n=4$ to $n=2$ in the $He^{+}$ spectrum is given by the Rydberg formula: $\bar{
u} = R Z^{2} ( \frac{1}{n_{

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$2 \rightarrow 1$

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(D) The wave number for the transition of $n=4$ to $n=2$ in the $He^{+}$ spectrum is given by the Rydberg formula: $\bar{
u} = R Z^{2} ( \frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}} )$.For $He^{+}$,$Z=2$,$n_{1}=2$,and $n_{2}=4$:$\bar{
u} = R(2)^{2} ( \frac{1}{2^{2}} - \frac{1}{4^{2}} ) = 4R ( \frac{1}{4} - \frac{1}{16} ) = 4R ( \frac{3}{16} ) = \frac{3R}{4}$.Thus,the wavelength $\lambda = \frac{4}{3R}$.For the hydrogen atom $(Z=1)$,we seek a transition $n_{2} ightarrow n_{1}$ such that $\bar{
u} = R(1)^{2} ( \frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}} ) = \frac{3R}{4}$.This simplifies to $( \frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}} ) = \frac{3}{4}$.Testing values,if $n_{1}=1$ and $n_{2}=2$,then $( \frac{1}{1^{2}} - \frac{1}{2^{2}} ) = 1 - \frac{1}{4} = \frac{3}{4}$.Therefore,the transition $n=2 ightarrow n=1$ in hydrogen has the same wavelength.

What transition in the hydrogen spectrum has the same wavelength as the Balmer transition,$n=4$ to $n=2$,of the $He^{+}$ spectrum?

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