Question

What is the wave number (Units cm$^{-1}$) of the longest wavelength transition in the Balmer series of the Hydrogen spectrum? $ Z = 1 $ for $ H $

• A. 27419.6
• B. 15233.3
• C. 39605.6
• D. 1354

Hint:

For transitions in the hydrogen atom, the Rydberg formula helps in calculating the wave number or wavelength for different series. The Balmer series corresponds to transitions where \( n_2 > n_1 \), and the longest wavelength corresponds to the transition from \( n_2 = 2 \) to \( n_1 = 3 \).

Answer 1

The Correct Option is B

The wave number \( \tilde{\nu} \) (in cm\(^{-1}\)) of any transition in the hydrogen atom can be found using the Rydberg formula for the Balmer series: \[ \tilde{\nu} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \] where: - \( R_H \) is the Rydberg constant for hydrogen (\( R_H = 1.097 \times 10^5 \, \text{cm}^{-1} \)), - \( n_1 \) is the lower energy level for the transition, - \( n_2 \) is the higher energy level for the transition. For the longest wavelength transition in the Balmer series, the transition occurs from \( n_2 = 2 \) to \( n_1 = 3 \). Thus, we can calculate the wave number: \[ \tilde{\nu} = 1.097 \times 10^5 \left( \frac{1}{2^2} - \frac{1}{3^2} \right) \] \[ \tilde{\nu} = 1.097 \times 10^5 \left( \frac{1}{4} - \frac{1}{9} \right) \] \[ \tilde{\nu} = 1.097 \times 10^5 \left( \frac{5}{36} \right) \] \[ \tilde{\nu} = 15233.3 \, \text{cm}^{-1} \] Therefore, the wave number for the longest wavelength transition in the Balmer series of the hydrogen spectrum is \( 15233.3 \, \text{cm}^{-1} \).