What is the ratio of the wave number of the f | vedclass

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

Vedclass · Accepted Answer

$5$ : $0.81$

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(B) The wave number is given by the Rydberg formula: $\bar{
u} = R_H Z^2 (\frac{1}{n_1^2} - \frac{1}{n_2^2})$.For the $H$ atom, $Z = 1$.For the first line of the Balmer series, $n_1 = 2$ and $n_2 = 3$. Thus, $\bar{
u}_B = R_H (\frac{1}{2^2} - \frac{1}{3^2}) = R_H (\frac{1}{4} - \frac{1}{9}) = R_H (\frac{9-4}{36}) = R_H (\frac{5}{36})$.For the first line of the Brackett series, $n_1 = 4$ and $n_2 = 5$. Thus, $\bar{
u}_{Br} = R_H (\frac{1}{4^2} - \frac{1}{5^2}) = R_H (\frac{1}{16} - \frac{1}{25}) = R_H (\frac{25-16}{400}) = R_H (\frac{9}{400})$.The ratio is $\frac{\bar{
u}_B}{\bar{
u}_{Br}} = \frac{R_H (5/36)}{R_H (9/400)} = \frac{5}{36} 	imes \frac{400}{9} = \frac{5 	imes 100}{9 	imes 9} = \frac{500}{81}$.Calculating the value, $\frac{500}{81} \approx 6.1728$.Comparing with option $B$, $\frac{5}{0.81} = \frac{500}{81} \approx 6.1728$. Therefore, the correct ratio is $5 : 0.81$.

What is the ratio of the wave number of the first line (lowest energy line) of the Balmer series of the $H$ atomic spectrum to the first line of its Brackett series?

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