Question

Energy levels \(\mathrm{A}, \mathrm{B}, \mathrm{C}\) of a certain atom corresponding values of energy i.e. \(E_{A}

Short answer

The correct relationship between the wavelengths is: \( \lambda_{3} = \frac{\lambda_{1} \lambda_{2}}{\lambda_{1} + \lambda_{2}} \).

Step-by-step solution

Write down the energy differences

The energy differences for the transitions are: - ΔE₁: from C to B: EB - EC - ΔE₂: from B to A: EA - EB - ΔE₃: from C to A: EA - EC

Relate energy differences to corresponding wavelengths

Using the Planck-Einstein relation, we can write the energy differences in terms of the corresponding wavelengths: - ΔE₁ = h * (c/λ₁) - ΔE₂ = h * (c/λ₂) - ΔE₃ = h * (c/λ₃)

Use energy conservation to find the relationship

Since energy is conserved during these transitions, we can write: ΔE₁ + ΔE₂ = ΔE₃ Substituting the energy differences from Step 2, we get: \( h\frac{c}{\lambda_{1}} + h\frac{c}{\lambda_{2}} = h\frac{c}{\lambda_{3}} \)

Simplify the equation and isolate λ₃

Dividing both sides by Planck's constant (h) and the speed of light (c), we simplify the equation to: \( \frac{1}{\lambda_{1}} + \frac{1}{\lambda_{2}} = \frac{1}{\lambda_{3}} \) Taking the reciprocal of all terms, we obtain the relationship between the wavelengths: \( \lambda_{3} = \frac{\lambda_{1} \lambda_{2}}{\lambda_{1} + \lambda_{2}} \) Comparing this result with the given options, we find that the correct answer is: (C) \( \lambda_{3} = \left[\left(\lambda_{1} \lambda_{2}\right) / \left(\lambda_{1}+\lambda_{2}\right)\right] \)