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(D) From the energy level diagram,the energy difference for the transition from $C$ to $A$ is the sum of the energy differences for the transitions fr

Vedclass · Accepted Answer

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

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(D) From the energy level diagram,the energy difference for the transition from $C$ to $A$ is the sum of the energy differences for the transitions from $C$ to $B$ and $B$ to $A$.$(E_C - E_A) = (E_C - E_B) + (E_B - E_A)$Using the relation $E = \frac{hc}{\lambda}$,we can write:$\frac{hc}{\lambda_3} = \frac{hc}{\lambda_1} + \frac{hc}{\lambda_2}$Dividing both sides by $hc$:$\frac{1}{\lambda_3} = \frac{1}{\lambda_1} + \frac{1}{\lambda_2}$$\frac{1}{\lambda_3} = \frac{\lambda_2 + \lambda_1}{\lambda_1 \lambda_2}$Therefore,$\lambda_3 = \frac{\lambda_1 \lambda_2}{\lambda_1 + \lambda_2}$.

Energy levels $A$,$B$ and $C$ of a certain atom correspond to increasing values of energy,i.e.,$E_A < E_B < E_C$. If $\lambda_1$,$\lambda_2$,and $\lambda_3$ are the wavelengths of the photons corresponding to the transitions shown in the figure,then:

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