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

The longest wavelength of spectral lines in the Balmer series is determined by examining the energy levels of hydrogen and the relevant electron transitions.

Step 1: Lyman Series Analysis
The energy-wavelength relationship for the Lyman series is expressed as:

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

where \( n_1 = 1 \) and \( n_2 \) varies. The shortest wavelength observed in the Lyman series is:

\[ \lambda_{\text{Lyman}} = 915 \, \text{\AA}. \]

Step 2: Balmer Series Calculation
For the Balmer series, the initial and final energy levels are:
- \( n_1 = 2 \)
- \( n_2 = 3 \) for the longest wavelength transition.

The energy difference for this transition is:

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

This simplifies to:

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

Further simplification yields:

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

Step 3: Wavelength Relationship Determination
Utilizing the data from the Lyman series, the wavelength for the transition in the Balmer series can be related as follows:

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

Substituting the known value for \( \lambda_{\text{Lyman}} \):

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

Consequently, the longest wavelength spectral line in the Balmer series measures \( 6588 \, \text{\AA} \).