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(C) 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 the

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$n = 4$ to $n = 2$

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(C) 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 the first Lyman transition of hydrogen $(H)$,$Z_H = 1$,$n_1 = 1$,and $n_2 = 2$. Thus,$\frac{1}{\lambda_H} = R(1)^2 \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = R \left( 1 - \frac{1}{4} \right) = \frac{3R}{4}$.For $He^+$,$Z_{He} = 2$. We want to find $n_1$ and $n_2$ such that $\frac{1}{\lambda_{He}} = \frac{1}{\lambda_H} = \frac{3R}{4}$.Substituting the values: $R(2)^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) = \frac{3R}{4}$.$4R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) = \frac{3R}{4} \implies \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) = \frac{3}{16}$.Comparing this to $\left( \frac{1}{2^2} - \frac{1}{4^2} \right) = \left( \frac{1}{4} - \frac{1}{16} \right) = \frac{4-1}{16} = \frac{3}{16}$.Thus,the transition is from $n = 4$ to $n = 2$.

Neglecting the reduced mass effect,which optical transition in the $He^+$ spectrum has the same wavelength as the first Lyman transition of hydrogen ($n = 2$ to $n = 1$)?

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