Question

Three energy levels of hydrogen atom and the corresponding wavelength of the emitted radiation due to different electron transition are as shown. Then

• A. \(λ_1=\frac{λ_2λ_3}{λ_2+λ_3}\)
• B. λ₂ = λ₁ + λ₃
• C. \(λ_2=\frac{λ_1λ_3}{λ_1+λ_3}\)
• D. \(λ_3=\frac{λ_1λ_2}{λ_1+λ_2}\)

Answer 1

The Correct Option is C

Step 1: Identify the given energy levels and transitions clearly: 
Three energy levels (E₁ < E₂ < E₃) of a hydrogen atom are given. Wavelengths of radiation emitted due to transitions between these levels are λ₁, λ₂, and λ₃:

• Transition from level E₃ → E₂ emits radiation λ₁.
• Transition from level E₂ → E₁ emits radiation λ₃.
• Transition from level E₃ → E₁ emits radiation λ₂.

Step 2: Use the energy-wavelength relationship clearly:
The energy difference (ΔE) between two levels is related to the wavelength (λ) of emitted radiation by:

\[ ΔE = \frac{hc}{λ} \]

where \(h\) = Planck’s constant, \(c\) = speed of light.

Step 3: Express the energy differences for each transition clearly:

• For transition E₃ → E₂: \[ E_3 - E_2 = \frac{hc}{λ_1} \]
• For transition E₂ → E₁: \[ E_2 - E_1 = \frac{hc}{λ_3} \]
• For transition E₃ → E₁: \[ E_3 - E_1 = \frac{hc}{λ_2} \]

Step 4: Relate the above energy differences:
Clearly, energy levels satisfy:

\[ (E_3 - E_1) = (E_3 - E_2) + (E_2 - E_1) \]

Substitute wavelength terms:

\[ \frac{hc}{λ_2} = \frac{hc}{λ_1} + \frac{hc}{λ_3} \]

Canceling out \(hc\), we get clearly:

\[ \frac{1}{λ_2} = \frac{1}{λ_1} + \frac{1}{λ_3} \]

Step 5: Solve this clearly to find λ₂ explicitly:

\[ λ_2 = \frac{λ_1λ_3}{λ_1 + λ_3} \]

Final Conclusion:
Thus, the correct relation is clearly:

\[ λ_2 = \frac{λ_1λ_3}{λ_1 + λ_3} \]