The shortest wavelength of hydrogen atom in L | vedclass

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(B) For $H$ atom in Lyman series,the shortest wavelength corresponds to the transition from $n_2 = \infty$ to $n_1 = 1$:$\frac{1}{\lambda} = R_H 	ime

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$\frac{9 \lambda}{5}$

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(B) For $H$ atom in Lyman series,the shortest wavelength corresponds to the transition from $n_2 = \infty$ to $n_1 = 1$:$\frac{1}{\lambda} = R_H 	imes 1^2 	imes (\frac{1}{1^2} - \frac{1}{\infty^2}) = R_H$  $(1)$For $He^{+}$ ion in Balmer series,the longest wavelength corresponds to the transition from $n_2 = 3$ to $n_1 = 2$:$\frac{1}{\lambda_{He^{+}}} = R_H 	imes 2^2 	imes (\frac{1}{2^2} - \frac{1}{3^2}) = R_H 	imes 4 	imes (\frac{1}{4} - \frac{1}{9}) = R_H 	imes 4 	imes \frac{5}{36} = R_H 	imes \frac{5}{9}$  $(2)$Dividing $(1)$ by $(2)$:$\frac{\lambda_{He^{+}}}{\lambda} = \frac{R_H}{R_H 	imes \frac{5}{9}} = \frac{9}{5}$Therefore,$\lambda_{He^{+}} = \frac{9 \lambda}{5}$.

The shortest wavelength of hydrogen atom in Lyman series is $\lambda$. The longest wavelength in Balmer series of $He^{+}$ is

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