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:

Answers (1)

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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