Question

A spectral line of Lyman series of H-atom has a frequency of \(2.466 \times 10^{15} \, {s}^{-1}\). What is the transition responsible for this spectral line? (Rydberg constant \(R = 1.096 \times 10^7 \, {m}^{-1}\))

• A. \(n_2 = 2 \to n_1 = 1\)
• B. \(n_2 = 3 \to n_1 = 1\)
• C. \(n_2 = 4 \to n_1 = 2\)
• D. \(n_2 = 5 \to n_1 = 1\)

Hint:

The Lyman series corresponds to electronic transitions where the final energy level is \(n_1=1\). Use the Rydberg formula and compare given frequency with calculated values to identify transitions.

Answer 1

The Correct Option is A

The frequency of spectral lines in the hydrogen atom is given by the Rydberg formula: \[ \nu = R c \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \] where \(\nu\) = frequency of spectral line, \(R\) = Rydberg constant \(= 1.096 \times 10^7 \, {m}^{-1}\), \(c\) = speed of light \(= 3 \times 10^8 \, {m/s}\), \(n_1, n_2\) = principal quantum numbers with \(n_2>n_1\). 
Given \(\nu = 2.466 \times 10^{15} \, {s}^{-1}\), Calculate: \[ \frac{\nu}{R c} = \frac{2.466 \times 10^{15}}{1.096 \times 10^{7} \times 3 \times 10^{8}} = \frac{2.466 \times 10^{15}}{3.288 \times 10^{15}} \approx 0.75 \] Therefore, \[ \frac{1}{n_1^2} - \frac{1}{n_2^2} = 0.75 \] Check possible transitions for Lyman series (\(n_1 = 1\)): - For \(n_2 = 2\): \[ 1 - \frac{1}{2^2} = 1 - \frac{1}{4} = \frac{3}{4} = 0.75 \] - For \(n_2 = 3\): \[ 1 - \frac{1}{9} = \frac{8}{9} \approx 0.888 \] - For \(n_2 = 4\): \[ 1 - \frac{1}{16} = \frac{15}{16} = 0.9375 \] - For \(n_2 = 5\): \[ 1 - \frac{1}{25} = \frac{24}{25} = 0.96 \] Only \(n_2 = 2 \to n_1 = 1\) satisfies the equation.