Question: Energy levels A, B and C of an atom correspond to increasing values of energy i.e. $E_A < E_B < E_C$. Let $\lambda_1$, $\lambda_2$ and $\lambda_3$ be the wavelengths of radiation corresponding to the transitions C to B, B to A and C to A, respectively. The correct relation between $\lambda_1$, $\lambda_2$ and $\lambda_3$ is:
Step 1 (Formula / Governing Law & Given Data): [1/3 Mark]
According to the Bohr's postulate, the energy of the emitted photon during a transition between two energy levels $E_2$ and $E_1$ is given by:
$$\Delta E = \frac{hc}{\lambda} = E_2 - E_1$$Given transitions:
- C to B: $\frac{hc}{\lambda_1} = E_C - E_B$
- B to A: $\frac{hc}{\lambda_2} = E_B - E_A$
- C to A: $\frac{hc}{\lambda_3} = E_C - E_A$
Step 2 (Step-by-Step Mathematical Working): [1/3 Mark]
Observe that the energy difference for the transition C to A is the sum of the energy differences for transitions C to B and B to A:
$$(E_C - E_A) = (E_C - E_B) + (E_B - E_A)$$Substituting the expressions derived in Step 1 into this equation:
$$\frac{hc}{\lambda_3} = \frac{hc}{\lambda_1} + \frac{hc}{\lambda_2}$$Dividing both sides by $hc$:
$$\frac{1}{\lambda_3} = \frac{1}{\lambda_1} + \frac{1}{\lambda_2}$$Step 3 (Final Concluded Answer): [1/3 Mark]
Rearranging the terms, we get:
$$\frac{1}{\lambda_1} + \frac{1}{\lambda_2} = \frac{1}{\lambda_3}$$Correct Option: (B)
💡 EL10 Concept Insight: Think of energy levels like steps on a ladder; the total energy released jumping from the top step (C) to the bottom (A) is exactly the same as jumping from the top (C) to the middle (B) and then from the middle (B) to the bottom (A). Since energy is inversely proportional to wavelength ($E \propto 1/\lambda$), adding the energies means adding their reciprocal wavelengths.