What transition in He^{+} ion shall have the | vedclass

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Question

(A) The wave number $\bar{
u}$ for a hydrogen-like species is given by the Rydberg formula: $\bar{
u} = R Z^2 (\frac{1}{n_1^2} - \frac{1}{n_2^2})$.F

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$6 \rightarrow 4$

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(A) The wave number $\bar{
u}$ for a hydrogen-like species is given by the Rydberg formula: $\bar{
u} = R Z^2 (\frac{1}{n_1^2} - \frac{1}{n_2^2})$.For the first line of the Balmer series in the hydrogen atom $(Z=1)$,the transition is from $n_2=3$ to $n_1=2$: $\bar{
u} = R (1)^2 (\frac{1}{2^2} - \frac{1}{3^2}) = R (\frac{1}{4} - \frac{1}{9}) = R (\frac{5}{36})$.For the $He^{+}$ ion $(Z=2)$,we need to find the transition $n_2 ightarrow n_1$ such that $\bar{
u} = R (2)^2 (\frac{1}{n_1^2} - \frac{1}{n_2^2}) = R \frac{5}{36}$.Simplifying,$4 (\frac{1}{n_1^2} - \frac{1}{n_2^2}) = \frac{5}{36}$,which gives $(\frac{1}{n_1^2} - \frac{1}{n_2^2}) = \frac{5}{144}$.Testing the options,for $n_2=4$ and $n_1=2$: $(\frac{1}{4} - \frac{1}{16}) = \frac{4-1}{16} = \frac{3}{16} 
eq \frac{5}{144}$.For $n_2=6$ and $n_1=4$: $(\frac{1}{16} - \frac{1}{36}) = \frac{9-4}{144} = \frac{5}{144}$.Thus,the transition $6 ightarrow 4$ in $He^{+}$ matches the wave number.

What transition in $He^{+}$ ion shall have the same wave number as the first line in Balmer series of hydrogen atom?

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