The wave number of the first Lyman transition | vedclass

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(C) The wave number $\bar{
u}$ is given by the Rydberg formula: $\bar{
u} = R_H Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} ight)$.For the first

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$He^{+}$

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(C) The wave number $\bar{
u}$ is given by the Rydberg formula: $\bar{
u} = R_H Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} ight)$.For the first Lyman transition in $H$ atom $(Z=1)$: $n_1=1, n_2=2$. So,$\bar{
u}_1 = R_H (1)^2 \left( \frac{1}{1^2} - \frac{1}{2^2} ight) = R_H \left( 1 - \frac{1}{4} ight) = \frac{3}{4} R_H$.For the second Balmer transition in a hydrogen-like ion $(Z)$: $n_1=2, n_2=4$. So,$\bar{
u}_2 = R_H Z^2 \left( \frac{1}{2^2} - \frac{1}{4^2} ight) = R_H Z^2 \left( \frac{1}{4} - \frac{1}{16} ight) = R_H Z^2 \left( \frac{3}{16} ight)$.Equating the two: $\frac{3}{4} R_H = R_H Z^2 \left( \frac{3}{16} ight)$.$\frac{3}{4} = Z^2 \left( \frac{3}{16} ight) \implies Z^2 = \frac{3}{4} 	imes \frac{16}{3} = 4$.Thus,$Z = 2$. The ion with $Z=2$ is $He^{+}$.

The wave number of the first Lyman transition in $H$ atom spectrum is equal to the wave number of the second Balmer transition in the spectrum of

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