If 917 $$\mathring{\text{A}}$$ be the lowest wavelength of Lyman series then the lowest wavelength of Balmer series will be _______ $$\mathring{\text{A}}$$.

We need to find the lowest wavelength of the Balmer series, given that the lowest wavelength of the Lyman series is 917 angstroms.

The wavelength of spectral lines in the hydrogen spectrum is given by:

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

where $$R$$ is the Rydberg constant, $$n_1$$ is the lower energy level, and $$n_2$$ is the upper energy level ($$n_2 > n_1$$).

Find the lowest wavelength (series limit) for each series.

The lowest wavelength in any series occurs when $$n_2 \to \infty$$, giving the series limit.

Lyman series ($$n_1 = 1$$): The series limit is:

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

So $$\lambda_L = \frac{1}{R} = 917$$ angstroms (given).

Balmer series ($$n_1 = 2$$): The series limit is:

$$\frac{1}{\lambda_B} = R\left(\frac{1}{2^2} - \frac{1}{\infty}\right) = \frac{R}{4}$$

So $$\lambda_B = \frac{4}{R}$$.

Since $$\lambda_L = \frac{1}{R}$$, we have $$R = \frac{1}{\lambda_L}$$. Substituting:

$$\lambda_B = \frac{4}{R} = 4\lambda_L = 4 \times 917 = 3668 \text{ angstroms}$$

The lowest wavelength of the Balmer series is 3668 angstroms.