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Electron in hydrogen atom first jumps from third excited state to second excited state and then from second excited to the first excited state. The ratio of the wavelengths $\lambda_1 : \lambda_2$ emitted in the two cases is
A
$\frac{20}{7}$
B
$\frac{7}{5}$
C
$\frac{27}{20}$
D
$\frac{27}{5}$
Detailed Solution
Third excited state is n = 4, second excited state is n = 3 and first excited state is n = 2.
$\frac{1}{\lambda_1} = R\left(\frac{1}{3^2} - \frac{1}{4^2}\right) = R\left(\frac{1}{9} - \frac{1}{16}\right) = \frac{7R}{144}$
$\frac{1}{\lambda_2} = R\left(\frac{1}{2^2} - \frac{1}{3^2}\right) = R\left(\frac{1}{4} - \frac{1}{9}\right) = \frac{5R}{36}$
$\frac{\lambda_1}{\lambda_2} = \frac{5R/36}{7R/144} = \frac{20}{7}$
$\frac{1}{\lambda_1} = R\left(\frac{1}{3^2} - \frac{1}{4^2}\right) = R\left(\frac{1}{9} - \frac{1}{16}\right) = \frac{7R}{144}$
$\frac{1}{\lambda_2} = R\left(\frac{1}{2^2} - \frac{1}{3^2}\right) = R\left(\frac{1}{4} - \frac{1}{9}\right) = \frac{5R}{36}$
$\frac{\lambda_1}{\lambda_2} = \frac{5R/36}{7R/144} = \frac{20}{7}$