Given the value of Rydberg constant is 10^7 , | vedclass

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(B) The wave number $\bar{
u}$ for the hydrogen spectrum is given by the Rydberg formula: $\bar{
u} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} i

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$0.25 	imes 10^7 \, m^{-1}$

Vedclass · Answer

(B) The wave number $\bar{
u}$ for the hydrogen spectrum is given by the Rydberg formula: $\bar{
u} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} ight)$.For the Balmer series,the transition ends at $n_1 = 2$.The last line of the Balmer series corresponds to the transition from $n_2 = \infty$ to $n_1 = 2$.Given $R = 10^7 \, m^{-1}$,we substitute the values:$\bar{
u} = 10^7 \left( \frac{1}{2^2} - \frac{1}{\infty^2} ight) = 10^7 \left( \frac{1}{4} - 0 ight) = \frac{10^7}{4} = 0.25 	imes 10^7 \, m^{-1}$.

Given the value of Rydberg constant is $10^7 \, m^{-1}$,the wave number of the last line of the Balmer series in the hydrogen spectrum will be:

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