The shortest wavelength in the Balmer series of a hydrogen atom is
(Given: $R=1.1 × 10^7 m^{-1}$)
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 3637 Å
The wavelength of spectral lines is given by Rydberg’s formula:
$$\frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$$
For Balmer series: $n_1 = 2$, $n_2 = 3,4,5,\dots$
The shortest wavelength corresponds to $n_2 \to \infty$:
$$\frac{1}{\lambda_{\min}} = R \left( \frac{1}{2^2} - \frac{1}{\infty^2} \right) = R \cdot \frac{1}{4}$$
Hence,
$$\lambda_{\min} = \frac{4}{R}$$
Taking $R = 1.097 \times 10^7 \ \text{m}^{-1}$,
$$\lambda_{\min} = \frac{4}{1.097 \times 10^7} \approx 3.64 \times 10^{-7} \ \text{m} = 364 \ \text{nm}$$
Answer: $\lambda_{\min} = 364 \ \text{nm}$