CBSE · Class 12 · Physics · AtomsCalculate the shortest wavelength of the spectral lines in the Balmer series of hydrogen atom. Given the value of Rydberg constant $R = 1.097 \times 10^7 \text{ m}^{-1}$.

To find the shortest wavelength of the spectral lines in the Balmer series of a hydrogen atom, we use the Rydberg formula for the wavelength $\lambda$ of emitted radiation:

$$\frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$$ \nFor the Balmer series, the lower energy level is fixed at $n_1 = 2$. \nTo obtain the shortest wavelength (which corresponds to the maximum energy transition), the upper energy level $n_2$ must be at infinity, i.e., $n_2 = \infty$. \nSubstituting the values into the formula:

$$\frac{1}{\lambda} = R \left( \frac{1}{2^2} - \frac{1}{\infty^2} \right)$$

$$\frac{1}{\lambda} = R \left( \frac{1}{4} - 0 \right)$$

$$\frac{1}{\lambda} = \frac{R}{4}$$

$$\lambda = \frac{4}{R}$$ \nNow, substitute the given value of the Rydberg constant $R = 1.097 \times 10^7 \text{ m}^{-1}$:

$$\lambda = \frac{4}{1.097 \times 10^7 \text{ m}^{-1}}$$

$$\lambda = \frac{4000}{1.097} \times 10^{-7} \text{ m}$$

$$\lambda \approx 3648 \times 10^{-10} \text{ m} = 364.8 \text{ nm}$$ \nThus, the shortest wavelength in the Balmer series is $3.648 \times 10^{-7} \text{ m}$ or $364.8 \text{ nm}$.