If Rydberg's constant is R,the longest wavele | vedclass

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(A) The wavelength $\lambda$ of radiation emitted in a hydrogen atom is given by the Rydberg formula: $\frac{1}{\lambda} = R \left( \frac{1}{n_1^2} -

Vedclass · Accepted Answer

$144$

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(A) The wavelength $\lambda$ of radiation emitted in a hydrogen atom is given by the Rydberg formula: $\frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$.For the Paschen series,the transition occurs to the $n_1 = 3$ energy level.The longest wavelength corresponds to the transition from the nearest energy level,which is $n_2 = 4$.Substituting these values: $\frac{1}{\lambda} = R \left( \frac{1}{3^2} - \frac{1}{4^2} \right) = R \left( \frac{1}{9} - \frac{1}{16} \right)$.$\frac{1}{\lambda} = R \left( \frac{16 - 9}{144} \right) = \frac{7R}{144}$.Therefore,$\lambda = \frac{144}{7R}$.Comparing this with $\frac{\alpha}{7R}$,we get $\alpha = 144$.

If Rydberg's constant is $R$,the longest wavelength of radiation in Paschen series will be $\frac{\alpha}{7 R}$,where $\alpha=$ . . . . . .

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