Question

Energy of the transition from \( n_h = 4 \) to \( n_l = 2 \) for hydrogen atom is \( E \times 10^3 \) cm\(^{-1}\).
Given: Rydberg constant for hydrogen: \( 1.097 \times 10^7 \, {m}^{-1} \)
Value of E is ............... (rounded off to two decimal places)

Hint:

To calculate the energy of a transition in the hydrogen atom, use the Rydberg formula, which accounts for the initial and final quantum numbers.

Answer 1

Correct Answer: 20.55

The energy of the transition in a hydrogen atom can be calculated using the Rydberg formula: \[ E = R_H \left( \frac{1}{n_l^2} - \frac{1}{n_h^2} \right) \] where \( R_H \) is the Rydberg constant for hydrogen, and \( n_l \) and \( n_h \) are the lower and higher quantum numbers, respectively. Substituting the given values: \[ E = (1.097 \times 10^7) \left( \frac{1}{2^2} - \frac{1}{4^2} \right) \] \[ E = (1.097 \times 10^7) \left( \frac{1}{4} - \frac{1}{16} \right) \] \[ E = (1.097 \times 10^7) \times \frac{3}{16} \] \[ E = 20.57 \times 10^3 \, {cm}^{-1} \] Thus, the correct value of \( E \) is \( 20.57 \times 10^3 \, {cm}^{-1} \).