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(A) The Rydberg formula for the wavelength $\lambda$ 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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(A) The Rydberg formula for the wavelength $\lambda$ 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 $n_1 = 2$ is:$\frac{1}{\lambda} = R (2)^2 \left( \frac{1}{2^2} - \frac{1}{4^2} \right) = 4R \left( \frac{1}{4} - \frac{1}{16} \right) = 4R \left( \frac{3}{16} \right) = R \left( \frac{3}{4} \right)$.For Hydrogen $(Z = 1)$,we want the transition from $n_2$ to $n_1$ such that:$\frac{1}{\lambda} = R (1)^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) = R \left( \frac{3}{4} \right)$.Comparing the two,we need $\frac{1}{n_1^2} - \frac{1}{n_2^2} = \frac{3}{4} = 1 - \frac{1}{4} = \frac{1}{1^2} - \frac{1}{2^2}$.Thus,the transition is from $n = 2$ to $n = 1$.

Which transition in the hydrogen spectrum would have the same wavelength as the Balmer type transition from $n = 4$ to $n = 2$ of $He^+$ spectrum?

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