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(D) The Rydberg formula for the wavelength of a transition is given by $\frac{1}{\lambda} = R_{H} Z^2 \left[ \frac{1}{n_1^2} - \frac{1}{n_2^2} \right]

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$n=2 \longrightarrow n=1$

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(D) The Rydberg formula for the wavelength of a transition is given by $\frac{1}{\lambda} = R_{H} Z^2 \left[ \frac{1}{n_1^2} - \frac{1}{n_2^2} \right]$.For the $He^{+}$ ion,$Z=2$. The transition from $n_2=4$ to $n_1=2$ gives:$\frac{1}{\lambda} = R_{H} (2)^2 \left[ \frac{1}{2^2} - \frac{1}{4^2} \right] = 4 R_{H} \left[ \frac{1}{4} - \frac{1}{16} \right] = 4 R_{H} \left( \frac{3}{16} \right) = \frac{3}{4} R_{H}$.For the hydrogen atom $(H)$,$Z=1$. We check the transitions:For option $(d)$,$n_2=2$ to $n_1=1$:$\frac{1}{\lambda} = R_{H} (1)^2 \left[ \frac{1}{1^2} - \frac{1}{2^2} \right] = R_{H} \left( 1 - \frac{1}{4} \right) = \frac{3}{4} R_{H}$.Since the values match,option $(d)$ is the correct answer.

Among the following,which transition in the hydrogen spectrum would have the same wavelength as the Balmer transition,$n=4$ to $n=2$ in the $He^{+}$ spectrum?

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