If the shortest wavelength of H atom in Lyman | vedclass

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(A) The Rydberg formula is given by $\frac{1}{\lambda} = R_{H} Z^2 \left[ \frac{1}{n_1^2} - \frac{1}{n_2^2} \right]$.For the Lyman series of $H$ atom

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$\frac{9x}{5}$

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(A) The Rydberg formula is given by $\frac{1}{\lambda} = R_{H} Z^2 \left[ \frac{1}{n_1^2} - \frac{1}{n_2^2} ight]$.For the Lyman series of $H$ atom $(Z=1)$,the shortest wavelength occurs at $n_1 = 1$ and $n_2 = \infty$.$\frac{1}{x} = R_{H} (1)^2 \left[ \frac{1}{1^2} - \frac{1}{\infty^2} ight] = R_{H} \implies R_{H} = \frac{1}{x}$.For the Balmer series of $He^{+}$ ion $(Z=2)$,the longest wavelength occurs at $n_1 = 2$ and $n_2 = 3$.$\frac{1}{\lambda_{\max}} = R_{H} (2)^2 \left[ \frac{1}{2^2} - \frac{1}{3^2} ight]$.Substituting $R_{H} = \frac{1}{x}$:$\frac{1}{\lambda_{\max}} = \frac{1}{x} 	imes 4 \left[ \frac{1}{4} - \frac{1}{9} ight] = \frac{4}{x} \left[ \frac{5}{36} ight] = \frac{5}{9x}$.Therefore,$\lambda_{\max} = \frac{9x}{5}$.

If the shortest wavelength of $H$ atom in Lyman series is $x$,then the longest wavelength in Balmer series of $He^{+}$ is

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