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(C) 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 the same tr

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$4000$

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(C) 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 the same transition,the term $(\frac{1}{n_1^2} - \frac{1}{n_2^2})$ is constant.Therefore,$\lambda \propto \frac{1}{Z^2}$.For $He^{+}$,$Z = 2$,so $\lambda_{He^+} \propto \frac{1}{2^2} = \frac{1}{4}$.For $H$,$Z = 1$,so $\lambda_{H} \propto \frac{1}{1^2} = 1$.Thus,$\frac{\lambda_{H}}{\lambda_{He^+}} = \frac{Z_{He^+}^2}{Z_{H}^2} = \frac{2^2}{1^2} = 4$.Given $\lambda_{He^+} = 100 \ nm = 1000 \ \mathring{A}$.Therefore,$\lambda_{H} = 4 	imes 1000 \ \mathring{A} = 4000 \ \mathring{A}$.

The wavelength of a particular electron transition for $He^{+}$ is $100 \ nm$. The wavelength (in $\mathring{A}$) of $H$ atom for the same transition is

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