When the electron in hydrogen atom jumps from the second orbit to the first orbit, the wavelength of the radiation emitted is $λ$ . When the electron jumps from the third to the first orbit, the wavelength of the radiation emitted is

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$\frac{27}{32} λ$

$\frac{9}{4} λ$

$\frac{32}{27} λ$

$\frac{4}{9} λ$

answer is C.

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Detailed Solution

$\begin{array}{l} \frac{1}{λ} = {Rz}^{2} (\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}}); ∴ \frac{1}{λ} \propto (\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}}) \\ ∴ \frac{λ_{1}}{λ_{2}} = \frac{\frac{1}{1^{2}} - \frac{1}{3^{2}}}{\frac{1}{1^{2}} - \frac{1}{2^{2}}} ∴ \frac{λ}{λ_{2}} = \frac{\frac{8}{3}}{\frac{3}{4}}; λ_{2} = \frac{27}{32} λ \end{array}$