Question

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.

Answer 1

To find the longest wavelength of the spectral lines in the Balmer series, we begin by using the relationship for the energy levels of hydrogen and the transitions involved.

Step 1: Lyman Series
For the Lyman series:

$\frac{hc}{\lambda} = -13.6 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \quad (\text{in eV}),$

where $n_1 = 1$ for the Lyman series and $n_2$ varies.
Given the shortest wavelength in the Lyman series:

$\lambda_{\text{Lyman}} = 915 \, \text{\AA}.$

Step 2: Balmer Series
For the Balmer series:
- $n_1 = 2$
- $n_2 = 3$ for the longest wavelength transition.

The energy difference for the transition is given by:

$\frac{hc}{\lambda_1} = -13.6 \left( \frac{1}{2^2} - \frac{1}{3^2} \right).$

Calculating:

$\frac{hc}{\lambda_1} = -13.6 \left( \frac{1}{4} - \frac{1}{9} \right).$

Simplifying:

$\frac{hc}{\lambda_1} = -13.6 \left( \frac{5}{36} \right).$

Step 3: Relating the Wavelengths
Using the given data for the Lyman series:

$\lambda_1 = \lambda_{\text{Lyman}} \times \frac{36}{5}.$

Substituting the given value:

$\lambda_1 = 915 \times \frac{36}{5} = 6588 \, \text{\AA}.$

Therefore, the longest wavelength of spectral lines in the Balmer series is $6588 \, \text{\AA}$.