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

Answer 2

Step 1: Given data.
Shortest wavelength in Lyman series (λ_L) = 915 Å
We need to find the longest wavelength in Balmer series (λ_B).

Step 2: Formula for wavelength of hydrogen spectral lines.
According to Rydberg formula:
\[ \frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \] where R = 1.097 × 10⁷ m⁻¹ (Rydberg constant).

Step 3: For Lyman series (n₁ = 1).
Shortest wavelength → n₂ = ∞
\[ \frac{1}{\lambda_L} = R \left( \frac{1}{1^2} - \frac{1}{\infty^2} \right) = R \] \[ \lambda_L = \frac{1}{R} \] Given λ_L = 915 Å = 9.15 × 10⁻⁸ m.
Hence, \[ R = \frac{1}{9.15 \times 10^{-8}} = 1.093 \times 10^7 \, \text{m}^{-1} \] which matches the standard Rydberg value.

Step 4: For Balmer series (n₁ = 2).
Longest wavelength → n₂ = 3.
\[ \frac{1}{\lambda_B} = R \left( \frac{1}{2^2} - \frac{1}{3^2} \right) \] \[ \frac{1}{\lambda_B} = R \left( \frac{1}{4} - \frac{1}{9} \right) = R \left( \frac{5}{36} \right) \] \[ \lambda_B = \frac{36}{5R} \] Substitute \( R = 1.093 \times 10^7 \, \text{m}^{-1} \):
\[ \lambda_B = \frac{36}{5 \times 1.093 \times 10^7} = 6.588 \times 10^{-7} \, \text{m} = 6588 \, \text{Å} \]

Step 5: Final Answer.
The longest wavelength of spectral lines in the Balmer series is:
\[ \boxed{6588 \, \text{Å}} \]

Final Answer: 6588 Å