Question:
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 $\mathrm{N}$-shell to the $\mathrm{L}$-shell, the wavelength of emitted radiation will be:
Correct Option: , 4
Solution:
(4) When electron jumps from $M \rightarrow L$ shell
$\frac{1}{\lambda}=K\left(\frac{1}{2^{2}}-\frac{1}{3^{2}}\right)=\frac{K \times 5}{36}$ ...(1)
When eletron jumps from $\mathrm{N} \rightarrow \mathrm{L}$ shell
$\frac{1}{\lambda^{\prime}}=\mathrm{K}\left(\frac{1}{2^{2}}-\frac{1}{4^{2}}\right)=\frac{\mathrm{K} \times 3}{16}$ ...(2)
solving equation (i) and (ii) we get
$\lambda^{\prime}=\frac{20}{27} \lambda$