Which transition in a H atom has the same wav | vedclass

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(B) The Rydberg formula for the wavelength of a transition is given by $\frac{1}{\lambda} = R Z^2 (\frac{1}{n_1^2} - \frac{1}{n_2^2})$.For $He^{+}$ io

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$n = 2$ to $n = 1$

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(B) The Rydberg formula for the wavelength of a transition is given by $\frac{1}{\lambda} = R Z^2 (\frac{1}{n_1^2} - \frac{1}{n_2^2})$.For $He^{+}$ ion $(Z = 2)$,the transition is from $n_2 = 4$ to $n_1 = 2$:$\frac{1}{\lambda} = R (2)^2 (\frac{1}{2^2} - \frac{1}{4^2}) = R (4) (\frac{1}{4} - \frac{1}{16}) = R (4) (\frac{3}{16}) = R (\frac{3}{4})$.For $H$ atom $(Z = 1)$,we want the same wavelength $\lambda$ for a transition from $n_2$ to $n_1$:$\frac{1}{\lambda} = R (1)^2 (\frac{1}{n_1^2} - \frac{1}{n_2^2}) = R (\frac{3}{4})$.Comparing the two,we get $(\frac{1}{n_1^2} - \frac{1}{n_2^2}) = \frac{3}{4} = (1 - \frac{1}{4}) = (\frac{1}{1^2} - \frac{1}{2^2})$.Thus,the transition is from $n_2 = 2$ to $n_1 = 1$.

Which transition in a $H$ atom has the same wavelength as the transition from $n = 4$ to $n = 2$ in a $He^{+}$ ion?

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