The shortest wavelength of the spectral lines in the Lyman series of hydrogen spectrum is $$915$$ Å. The longest wavelength of spectral lines in the Balmer series will be ______ Å.

We are given that the shortest wavelength of the Lyman series is 915 Angstrom, and we need to find the longest wavelength of the Balmer series.

Recall the Rydberg formula for hydrogen spectral lines

The wavelength of spectral lines in hydrogen 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.

Find the Rydberg constant from the Lyman series data

The Lyman series corresponds to transitions where $$n_1 = 1$$. The shortest wavelength (series limit) occurs when $$n_2 \to \infty$$:

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

Given $$\lambda_{\infty} = 915$$ Angstrom:

$$R = \frac{1}{915} \text{ per Angstrom}$$

Calculate the longest wavelength of the Balmer series

The Balmer series corresponds to transitions where $$n_1 = 2$$. The longest wavelength (smallest energy transition) occurs when $$n_2 = 3$$ (the closest upper level):

$$\frac{1}{\lambda} = R\left(\frac{1}{2^2} - \frac{1}{3^2}\right) = R\left(\frac{1}{4} - \frac{1}{9}\right)$$

Finding the common denominator:

$$\frac{1}{4} - \frac{1}{9} = \frac{9 - 4}{36} = \frac{5}{36}$$

Therefore:

$$\frac{1}{\lambda} = R \times \frac{5}{36} = \frac{1}{915} \times \frac{5}{36}$$

$$\lambda = \frac{36}{5R} = \frac{36 \times 915}{5} = \frac{32940}{5} = 6588 \text{ Angstrom}$$

The answer is 6588 Angstrom.