Q. For a certain atom, there are energy levels A, B, C corresponds to energy values $E_{A} < E_{B} < E_{C}$ . Choose the correct option if $\lambda _{1}, \, \lambda _{2}, \, \lambda _{3}$ are the wavelength of radiations corresponding to the transition from C to B, B to A and C to A respectively.

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A

$\lambda _{3}=\lambda _{1}+\lambda _{2}$

B

$\lambda _{3}=\frac{\lambda _{1} \lambda _{2}}{\lambda _{1} + \lambda _{2}}$

C

$\lambda _{1}+\lambda _{2}+\lambda _{3}=0$

D

$3\lambda _{2}=\lambda _{3}+2\lambda _{2}$

Solution:

The energy difference between levels C and A will be equal to the sum of energy difference between levels C & B and levels B & A.
$\text{E}_{\text{B}} - \text{E}_{\text{A}} = \frac{\text{h} \text{c}}{\text{\lambda }_{1}}$
$\text{E}_{\text{C}}-\text{E}_{\text{B}}=\frac{\text{h} \text{c}}{\text{\lambda }_{2}}$
$\text{E}_{\text{C}} - \text{E}_{\text{A}} = \frac{\text{h} \text{c}}{\text{\lambda }_{3}}$
Or $\frac{1}{\text{\lambda }_{3}}=\frac{1}{\text{\lambda }_{1}}+\frac{1}{\text{\lambda }_{2}}$
$\text{\lambda }_{3}=\frac{\text{\lambda }_{1} \text{\lambda }_{2}}{\text{\lambda }_{1} + \text{\lambda }_{2}}$

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