The energy levels of a certain atom are A, B, | vedclass

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(B) The energy of a photon emitted during a transition is given by $\Delta E = \frac{hc}{\lambda}$.For the transition $C 	o B$: $E_C - E_B = \frac{hc

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) The energy of a photon emitted during a transition is given by $\Delta E = \frac{hc}{\lambda}$.For the transition $C 	o B$: $E_C - E_B = \frac{hc}{\lambda_1} \quad ......(1)$For the transition $B 	o A$: $E_B - E_A = \frac{hc}{\lambda_2} \quad ......(2)$For the transition $C 	o A$: $E_C - E_A = \frac{hc}{\lambda_3} \quad ......(3)$From the energy level diagram,we can see that $(E_C - E_B) + (E_B - E_A) = E_C - E_A$.Substituting the values from equations $(1), (2),$ and $(3)$:$\frac{hc}{\lambda_1} + \frac{hc}{\lambda_2} = \frac{hc}{\lambda_3}$Dividing both sides by $hc$:$\frac{1}{\lambda_1} + \frac{1}{\lambda_2} = \frac{1}{\lambda_3}$$\frac{\lambda_2 + \lambda_1}{\lambda_1 \lambda_2} = \frac{1}{\lambda_3}$Therefore,$\lambda_3 = \frac{\lambda_1 \lambda_2}{\lambda_1 + \lambda_2}$.

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