The wavenumber of the first line (n_2=3) in t | vedclass

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(B) The Rydberg formula for wavenumber is $\bar{
u} = R_H Z^2 (\frac{1}{n_1^2} - \frac{1}{n_2^2})$.For the first line of the Balmer series of $H$ $(Z

Vedclass · Accepted Answer

$\frac{27 \bar{
u}_1}{5}$

Vedclass · Answer

(B) The Rydberg formula for wavenumber is $\bar{
u} = R_H Z^2 (\frac{1}{n_1^2} - \frac{1}{n_2^2})$.For the first line of the Balmer series of $H$ $(Z=1, n_1=2, n_2=3)$: $\bar{
u}_1 = R_H (1)^2 (\frac{1}{2^2} - \frac{1}{3^2}) = R_H (\frac{1}{4} - \frac{1}{9}) = R_H (\frac{5}{36})$.So,$R_H = \frac{36 \bar{
u}_1}{5}$.For the second line of the Balmer series of $He^{+}$ $(Z=2, n_1=2, n_2=4)$: $\bar{
u}_2 = R_H (2)^2 (\frac{1}{2^2} - \frac{1}{4^2}) = R_H (4) (\frac{1}{4} - \frac{1}{16}) = R_H (4) (\frac{3}{16}) = R_H (\frac{3}{4})$.Substituting $R_H = \frac{36 \bar{
u}_1}{5}$ into the expression for $\bar{
u}_2$: $\bar{
u}_2 = (\frac{36 \bar{
u}_1}{5}) 	imes (\frac{3}{4}) = \frac{9 	imes 3 \bar{
u}_1}{5} = \frac{27 \bar{
u}_1}{5}$.

The wavenumber of the first line $(n_2=3)$ in the Balmer series of hydrogen is $\bar{\nu}_1 \ cm^{-1}$. What is the wavenumber (in $cm^{-1}$) of the second line $(n_2=4)$ in the Balmer series of $He^{+}$?

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