The ratio of the lowest energy in terms of wa | vedclass

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Question

(A) The wave number $(\bar{
u})$ is given by the Rydberg formula: $\bar{
u} = R \cdot Z^2 \left( \frac{1}{n_{L}^2} - \frac{1}{n_{H}^2} ight)$.For

Vedclass · Accepted Answer

$5 : 27$

Vedclass · Answer

(A) The wave number $(\bar{
u})$ is given by the Rydberg formula: $\bar{
u} = R \cdot Z^2 \left( \frac{1}{n_{L}^2} - \frac{1}{n_{H}^2} ight)$.For hydrogen,$Z = 1$.For the lowest energy transition in the Balmer series,$n_{L} = 2$ and $n_{H} = 3$:$\bar{
u}_{Balmer} = R \left( \frac{1}{2^2} - \frac{1}{3^2} ight) = R \left( \frac{1}{4} - \frac{1}{9} ight) = R \left( \frac{9-4}{36} ight) = \frac{5}{36} R$.For the lowest energy transition in the Lyman series,$n_{L} = 1$ and $n_{H} = 2$:$\bar{
u}_{Lyman} = R \left( \frac{1}{1^2} - \frac{1}{2^2} ight) = R \left( 1 - \frac{1}{4} ight) = \frac{3}{4} R$.The ratio of the wave number of the Balmer series to the Lyman series is:$\frac{\bar{
u}_{Balmer}}{\bar{
u}_{Lyman}} = \frac{\frac{5}{36} R}{\frac{3}{4} R} = \frac{5}{36} 	imes \frac{4}{3} = \frac{5}{9 	imes 3} = \frac{5}{27}$.Thus,the ratio is $5 : 27$.

The ratio of the lowest energy in terms of wave numbers of the Balmer and Lyman series of lines of the atomic spectrum of hydrogen is

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