Question

Calculate the longest wavelength in hydrogen emission spectrum of Lyman series.
\([\text{R}_{\text{H}} = 109677\text{ cm}^{-1}]\)

• A. \(1.331 \times 10^{-5}\text{ cm}\)
• B. \(1.216 \times 10^{-5}\text{ cm}\)
• C. \(1.445 \times 10^{-5}\text{ cm}\)
• D. \(1.556 \times 10^{-5}\text{ cm}\)

Hint:

Longest wavelength → smallest energy gap → \(n_2 = 2\) in Lyman series.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The wavelength of light emitted during electron transitions in a hydrogen atom is given by the Rydberg formula. In the Lyman series, the electron falls to the ground state (\(n_1 = 1\)). The "longest wavelength" corresponds to the transition with the "minimum energy" change.
Step 2: Key Formula or Approach:
Rydberg equation: \[ \frac{1}{\lambda} = \text{R}_{\text{H}} \left( \frac{1}{\text{n}_1^2} - \frac{1}{\text{n}_2^2} \right) \] For Lyman longest wavelength: \(n_1 = 1\) and \(n_2 = 2\).
Step 3: Detailed Explanation:
\[ \frac{1}{\lambda} = 109677 \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = 109677 \left( 1 - 0.25 \right) \] \[ \frac{1}{\lambda} = 109677 \times 0.75 = 82257.75\text{ cm}^{-1} \] \[ \lambda = \frac{1}{82257.75} \approx 1.21569 \times 10^{-5}\text{ cm} \] Rounding to significant figures, we get \(1.216 \times 10^{-5}\text{ cm}\).
Step 4: Final Answer:
The wavelength is \(1.216 \times 10^{-5}\text{ cm}\).