Explanation

The wavelength of spectral lines in hydrogen atom is given by Rydberg formula

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

Statement A:

For Lyman series, $n_1 = 1$. Maximum wavelength corresponds to minimum energy transition, i.e. $n_2 = 2$.

$$ \frac{1}{\lambda_{\max}} = R\left(1-\frac{1}{4}\right) = \frac{3R}{4} $$

$$ \lambda_{\max} = \frac{4}{3R} $$

So Statement A is correct.

Statement B:

Balmer series corresponds to transitions ending at $n_1 = 2$ and lies in the visible region of the spectrum.

So Statement B is correct.

Statement C:

For Paschen series, $n_1 = 3$. Minimum wavelength corresponds to $n_2 = \infty$.

$$ \frac{1}{\lambda_{\min}} = R\left(\frac{1}{9}\right) $$

$$ \lambda_{\min} = \frac{9}{R} $$

So Statement C is correct.

Statement D:

For Lyman series minimum wavelength,

$$ \frac{1}{\lambda_{\min}} = R $$

$$ \lambda_{\min} = \frac{1}{R} $$

Hence $\dfrac{5}{4R}$ is incorrect.

Therefore, the correct choice is

$$ \boxed{\text{A, B and C Only}} $$

Related JEE Main Questions