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

Concept:
\[ \frac{1}{\lambda} = R_H \left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right) \] Step 1: For longest wavelength. In Lyman series: \[ n_1 = 1,\quad n_2 = 2 \]
Step 2: Substitute. \[ \frac{1}{\lambda} = 109677 \left(1 - \frac{1}{4}\right) = 109677 \times \frac{3}{4} \]
Step 3: Calculate. \[ \frac{1}{\lambda} = 82257.75 \Rightarrow \lambda = 1.216 \times 10^{-5}\text{ cm} \]
Step 4: Conclusion. Longest wavelength = \(1.216 \times 10^{-5}\text{ cm}\)