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

Correct Answer: 6588

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 \( 6588 \, \text{\AA} \).