The shortest wavelength of He^{+} ion in Balm | vedclass

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(B) The wavelength of $H$-like atoms is given by the Rydberg formula:$\frac{1}{\lambda} = R Z^2 \left[ \frac{1}{n_1^2} - \frac{1}{n_2^2} \right]$For t

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$\frac{16x}{7}$

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(B) The wavelength of $H$-like atoms is given by the Rydberg formula:$\frac{1}{\lambda} = R Z^2 \left[ \frac{1}{n_1^2} - \frac{1}{n_2^2} \right]$For the shortest wavelength $x$ of the Balmer series of $He^{+}$ $(Z=2)$: $n_1 = 2$ and $n_2 = \infty$.$\frac{1}{x} = R(2)^2 \left[ \frac{1}{2^2} - \frac{1}{\infty^2} \right] = 4R \left[ \frac{1}{4} \right] = R$$\Rightarrow x = \frac{1}{R}$For the longest wavelength $\lambda$ of the Paschen series of $Li^{2+}$ $(Z=3)$: $n_1 = 3$ and $n_2 = 4$.$\frac{1}{\lambda} = R(3)^2 \left[ \frac{1}{3^2} - \frac{1}{4^2} \right]$$\frac{1}{\lambda} = 9R \left[ \frac{1}{9} - \frac{1}{16} \right] = 9R \left[ \frac{16-9}{144} \right] = 9R \left[ \frac{7}{144} \right] = \frac{7R}{16}$$\Rightarrow \lambda = \frac{16}{7R} = \frac{16x}{7}$Hence,the correct option is $B$.

The shortest wavelength of $He^{+}$ ion in Balmer series is $x$. Then,the longest wavelength in the Paschen series of $Li^{2+}$ is:

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