The shortest wavelength in the Balmer series of a hydrogen atom is

(Given: $R=1.1 × 10^7 m^{-1}$)

3637 Å

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: $\lambda_{\min} = 364 \ \text{nm}$