Question

The wavelength of a photon emitted during a transition from \( n = 3 \) to \( n = 2 \) state in the H atom is .............. nm. (answer in integer).

Hint:

The wavelength of a photon emitted during a transition between energy levels in a hydrogen atom can be calculated using the Rydberg formula and the relationship between energy and wavelength.

Answer 1

Step 1: Formula for energy of the photon.
The energy of the photon emitted during the transition of an electron in a hydrogen atom from one energy level to another is given by the Rydberg formula: \[ E = - \frac{R_H}{n^2} \] Where \( R_H = 2.18 \times 10^{-18} \, \text{J} \) is the Rydberg energy constant and \( n \) is the principal quantum number.
The energy difference between the \( n = 3 \) and \( n = 2 \) states is: \[ \Delta E = \left( - \frac{R_H}{2^2} \right) - \left( - \frac{R_H}{3^2} \right) \] \[ \Delta E = R_H \left( \frac{1}{4} - \frac{1}{9} \right) = R_H \times \frac{5}{36} \] Substituting the value of \( R_H \): \[ \Delta E = 2.18 \times 10^{-18} \times \frac{5}{36} = 3.03 \times 10^{-19} \, \text{J} \] Step 2: Calculating the wavelength.
The energy of a photon is related to its wavelength \( \lambda \) by the equation: \[ E = \frac{h c}{\lambda} \] Where:
- \( h = 6.626 \times 10^{-34} \, \text{J s} \) is Planck’s constant
- \( c = 3 \times 10^8 \, \text{m/s} \) is the speed of light
Rearranging to solve for \( \lambda \): \[ \lambda = \frac{h c}{\Delta E} \] Substituting the values: \[ \lambda = \frac{(6.626 \times 10^{-34}) \times (3 \times 10^8)}{3.03 \times 10^{-19}} = 6.56 \times 10^{-7} \, \text{m} = 656 \, \text{nm} \] Final Answer: \[ \boxed{656 \, \text{nm}} \]