The shortest wavelength of H atom in the Lyma | vedclass

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(C) 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 shortest wavele

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$\frac{9 \lambda_{1}}{5}$

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(C) 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 shortest wavelength of the $H$ atom in the Lyman series $(n_{1} = 1)$,the transition is from $n_{2} = \infty$ to $n_{1} = 1$:$\frac{1}{\lambda_{1}} = R_{H} (1)^{2} \left( \frac{1}{1^{2}} - \frac{1}{\infty^{2}} ight) = R_{H}$ $\Rightarrow \lambda_{1} = \frac{1}{R_{H}}$.For the longest wavelength of the $He^{+}$ ion in the Balmer series $(n_{1} = 2)$,the transition is from $n_{2} = 3$ to $n_{1} = 2$:$\frac{1}{\lambda_{2}} = R_{H} (2)^{2} \left( \frac{1}{2^{2}} - \frac{1}{3^{2}} ight) = R_{H} 	imes 4 \left( \frac{1}{4} - \frac{1}{9} ight)$.$\frac{1}{\lambda_{2}} = 4 R_{H} \left( \frac{9-4}{36} ight) = 4 R_{H} \left( \frac{5}{36} ight) = \frac{5 R_{H}}{9}$.$\lambda_{2} = \frac{9}{5 R_{H}}$.Substituting $\frac{1}{R_{H}} = \lambda_{1}$,we get $\lambda_{2} = \frac{9 \lambda_{1}}{5}$.

The shortest wavelength of $H$ atom in the Lyman series is $\lambda_{1}$. The longest wavelength in the Balmer series of $He^{+}$ is:

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