Question

The wavelength of the radiation emitted, when an electron falls from infinity to stationary state 1, would be (Rydberg constant = 1.097 × 10⁷ m⁻¹)

• A. 406 nm
• B. 192 nm
• C. 91 nm
• D. 9.1 × 10⁻⁸ nm

Hint:

Key Exam Tip:
The Rydberg formula directly calculates the inverse of wavelength. Remember to convert your answer to the requested units (e.g., nm) and use the correct initial and final quantum numbers ($n_i, n_f$).

Answer 1

The Correct Option is C

The wavelength of emitted radiation during an electron transition in a hydrogen atom is given by the Rydberg formula:
$\frac{1}{\lambda} = R_H \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right)$
Here, $n_i = \infty$ (initial state is infinity), $n_f = 1$ (final state is the first stationary state), and $R_H = 1.097 \times 10^7 \text{ m}^{-1}$.
Substituting the values:
$\frac{1}{\lambda} = 1.097 \times 10^7 \text{ m}^{-1} \left( \frac{1}{1^2} - \frac{1}{\infty^2} \right)$
Since $1/\infty^2 = 0$:
$\frac{1}{\lambda} = 1.097 \times 10^7 \text{ m}^{-1} (1 - 0)$
$\frac{1}{\lambda} = 1.097 \times 10^7 \text{ m}^{-1}$
Solving for $\lambda$:
$\lambda = \frac{1}{1.097 \times 10^7 \text{ m}^{-1}} \approx 0.91158 \times 10^{-7} \text{ m}$
To convert meters to nanometers, multiply by $10^9$ (since $1 \text{ m} = 10^9 \text{ nm}$):
$\lambda = 0.91158 \times 10^{-7} \text{ m} \times 10^9 \text{ nm/m}$
$\lambda = 9.1158 \times 10^2 \text{ nm}$
$\lambda = 91.158 \text{ nm}$
The closest option is 91 nm.
Final Answer: \(\boxed{C}\)