In a hydrogen like atom, when an electron jumps from the M-shell to the L-shell,

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:

  1. (1) $\frac{27}{20} \lambda$

  2. (2) $\frac{16}{25} \lambda$

  3. (3) $\frac{25}{16} \lambda$

  4. (4) $\frac{20}{27} \lambda$


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

$\lambda^{\prime}=\frac{20}{27} \lambda$

 

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