The ratio of the shortest wavelength of Balmer series to the shortest wavelength of Lyman series for hydrogen atom is :

We need to find the ratio of the shortest wavelength of the Balmer series to the shortest wavelength of the Lyman series for hydrogen.

Recall the Rydberg formula:

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

For the Balmer series with $$n_1 = 2$$, the shortest wavelength (highest energy) occurs as $$n_2 \to \infty$$. Thus:

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

For the Lyman series with $$n_1 = 1$$, the shortest wavelength occurs as $$n_2 \to \infty$$. Hence:

$$ \frac{1}{\lambda_L} = R\left(1 - 0\right) = R \implies \lambda_L = \frac{1}{R} $$

Therefore the ratio is given by:

$$ \frac{\lambda_B}{\lambda_L} = \frac{4/R}{1/R} = \frac{4}{1} = 4:1 $$

The correct answer is Option (1): 4:1.