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(D) From the given energy level diagram,the energy differences for the transitions are:$E_2 - E_1 = \frac{hc}{\lambda_1}$  $(i)$$E_3 - E_2 = \frac{hc}

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$\lambda_2=\frac{\lambda_1 \lambda_3}{\lambda_1+\lambda_3}$

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(D) From the given energy level diagram,the energy differences for the transitions are:$E_2 - E_1 = \frac{hc}{\lambda_1}$  $(i)$$E_3 - E_2 = \frac{hc}{\lambda_3}$  (ii)$E_3 - E_1 = \frac{hc}{\lambda_2}$  (iii)Adding equations $(i)$ and (ii),we get:$(E_2 - E_1) + (E_3 - E_2) = \frac{hc}{\lambda_1} + \frac{hc}{\lambda_3}$$E_3 - E_1 = \frac{hc}{\lambda_1} + \frac{hc}{\lambda_3}$Substituting from equation (iii):$\frac{hc}{\lambda_2} = \frac{hc}{\lambda_1} + \frac{hc}{\lambda_3}$Dividing both sides by $hc$:$\frac{1}{\lambda_2} = \frac{1}{\lambda_1} + \frac{1}{\lambda_3}$$\frac{1}{\lambda_2} = \frac{\lambda_3 + \lambda_1}{\lambda_1 \lambda_3}$Therefore,$\lambda_2 = \frac{\lambda_1 \lambda_3}{\lambda_1 + \lambda_3}$.

Three energy levels of a hydrogen atom and the corresponding wavelengths of the emitted radiation due to different electron transitions are as shown. Then,

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