Question

The principal quantum number \( n \) corresponding to the excited state of \( He^+ \) ion, if on transition to the ground state two photons in succession with wavelengths \( 1026 \text{\AA} \) and \( 304 \text{\AA} \) are emitted: \((R = 1.097 \times 10^7 \text{ m}^{-1})\)

• A. \( 2 \)
• B. \( 3 \)
• C. \( 6 \)
• D. \( 4 \)

Hint:

For hydrogen-like atoms, the Rydberg formula helps determine transition wavelengths. The energy difference between levels is inversely proportional to the square of the principal quantum number.

Answer 1

The Correct Option is C

Step 1: Using the Rydberg Formula
For hydrogen-like species, the Rydberg formula for the wavelength of emitted photons is: \[ \frac{1}{\lambda} = RZ^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \] where: - \( R = 1.097 \times 10^7 \text{ m}^{-1} \) (Rydberg constant),
- \( Z = 2 \) for \( He^+ \),
- \( n_1 \) and \( n_2 \) are the quantum numbers of the final and initial states.
Step 2: Determining the Energy Levels
For the first transition, the given wavelength is \( \lambda_1 = 1026 \text{\AA} = 1026 \times 10^{-10} \text{ m} \), so: \[ \frac{1}{1026 \times 10^{-10}} = (1.097 \times 10^7) \times 4 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \] Solving for \( n_1 \) and \( n_2 \), this corresponds to the transition from \( n = 6 \) to \( n = 3 \).
For the second transition with \( \lambda_2 = 304 \text{\AA} \): \[ \frac{1}{304 \times 10^{-10}} = (1.097 \times 10^7) \times 4 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \] This corresponds to the transition from \( n = 3 \) to \( n = 1 \).
Thus, the electron was initially in the \( n = 6 \) state before transitioning to the ground state.
Final Answer: The principal quantum number \( n \) for the excited state of \( He^+ \) is \( 6 \), which matches option (3).