In a hydrogen-like atom, when an electron jumps from the M - shell to the L - shell, the wavelength of emitted radiation is $\lambda$. If an electron jumps from N - shell to the L - shell, the wavelength of emitted radiation will be:

Energy emitted due to transition of electron -

$\Delta E= Rhcz^{2}\left ( \frac{1}{n_{f}\, ^{2}}-\frac{1}{n_{i}\, ^{2}} \right )$

$\frac{1}{\lambda }= Rz^{2}\left ( \frac{-1}{n_{i}\, ^{2}}+\frac{1}{n_{f}\, ^{2}} \right )$

- wherein

$R= R hydberg\: constant$

$n_{i}= initial state \\n_{f}= final \: state$

$\left ( I \right )\: \: \: \: \: \: \: \: M\overset{\lambda _{1}}{\rightarrow}L$

$\left ( II \right )\: \: \: \: \: \: \: \: N\overset{\lambda _{2}}{\rightarrow}L$

$\lambda _{1}=\lambda$

For (I)

$\frac{1}{\lambda }=K\left ( \frac{1}{2^{2}}-\frac{1}{3^{2}} \right )=\frac{K\times 5}{36}\cdots (1)$

For (II)

$\frac{1}{\lambda_{2} }=K\left ( \frac{1}{2^{2}}-\frac{1}{4^{2}} \right )=\frac{3K}{16}\cdots (2)$

From (1) & (2)

$\frac{\lambda _{1}}{\lambda_{2} }=\frac{\left ( \frac{5K}{36} \right )}{\left ( \frac{3K}{16} \right )}$

$\lambda _{2}=\frac{20}{27}\lambda$

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