Explanation

For hydrogen atom, wavelength of spectral lines is given by Rydberg formula

$$ \frac{1}{\lambda} = R\left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right) $$

The smallest wavelength of the Lyman series corresponds to transition $n_2 \to \infty$ to $n_1 = 1$,

$$ \frac{1}{\lambda_{\text{Lyman(min)}}} = R $$

Given $\lambda_{\text{Lyman(min)}} = 91\ \text{nm}$,

$$ R = \frac{1}{91} $$

The largest wavelength of any series corresponds to transition $n_2 = n_1 + 1$.

For Balmer series, $n_1 = 2$,

$$ \frac{1}{\lambda_B} = R\left(\frac{1}{2^2} - \frac{1}{3^2}\right) = R\left(\frac{5}{36}\right) $$

$$ \lambda_B = \frac{36}{5R} = \frac{36}{5} \times 91 \approx 655\ \text{nm} $$

For Paschen series, $n_1 = 3$,

$$ \frac{1}{\lambda_P} = R\left(\frac{1}{3^2} - \frac{1}{4^2}\right) = R\left(\frac{7}{144}\right) $$

$$ \lambda_P = \frac{144}{7R} = \frac{144}{7} \times 91 \approx 1872\ \text{nm} $$

Difference between the largest wavelengths,

$$ \lambda_P - \lambda_B = 1872 - 655 \approx 1217\ \text{nm} $$

Hence, the required difference is

1217 nm

Related JEE Main Questions