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Energy levels A, B and C of a certain atom correspond to increasing values of energy, i.e. $E_A \lt E_B \lt E_C$. If $\lambda_1$, $\lambda_2$ and $\lambda_3$ are wavelengths of radiations corresponding to transitions C to B, B to A and C to A respectively, which of the following relations is correct -
A
$\lambda_3 = \lambda_1 + \lambda_2$
B
$\lambda_1 + \lambda_2 + \lambda_3 = 0$
C
$\lambda_3^2 = \lambda_1^2 + \lambda_2^2$
D
$\lambda_3 = \dfrac{\lambda_1\lambda_2}{\lambda_1 + \lambda_2}$
Detailed Solution
The energy of the photon emitted in a transition equals the energy difference between the levels: $\Delta E = \dfrac{hc}{\lambda}$.
C to B: $E_C - E_B = \dfrac{hc}{\lambda_1}$; B to A: $E_B - E_A = \dfrac{hc}{\lambda_2}$; C to A: $E_C - E_A = \dfrac{hc}{\lambda_3}$
Since $(E_C - E_B) + (E_B - E_A) = (E_C - E_A)$:
$\dfrac{hc}{\lambda_1} + \dfrac{hc}{\lambda_2} = \dfrac{hc}{\lambda_3}$
$\dfrac{1}{\lambda_3} = \dfrac{1}{\lambda_1} + \dfrac{1}{\lambda_2} = \dfrac{\lambda_1 + \lambda_2}{\lambda_1\lambda_2}$
$\Rightarrow \lambda_3 = \dfrac{\lambda_1\lambda_2}{\lambda_1 + \lambda_2}$
C to B: $E_C - E_B = \dfrac{hc}{\lambda_1}$; B to A: $E_B - E_A = \dfrac{hc}{\lambda_2}$; C to A: $E_C - E_A = \dfrac{hc}{\lambda_3}$
Since $(E_C - E_B) + (E_B - E_A) = (E_C - E_A)$:
$\dfrac{hc}{\lambda_1} + \dfrac{hc}{\lambda_2} = \dfrac{hc}{\lambda_3}$
$\dfrac{1}{\lambda_3} = \dfrac{1}{\lambda_1} + \dfrac{1}{\lambda_2} = \dfrac{\lambda_1 + \lambda_2}{\lambda_1\lambda_2}$
$\Rightarrow \lambda_3 = \dfrac{\lambda_1\lambda_2}{\lambda_1 + \lambda_2}$