Question

A monochromatic light is incident on a hydrogen sample in ground state. Hydrogen atoms absorb a fraction of light and subsequently emit radiation of six different wavelengths. The frequency of incident light is \( x \times 10^{15} \, \text{Hz} \). The value of \( x \) is_______

Hint:

The frequency of the incident light can be determined using the energy differences between electron energy levels in the hydrogen atom.

Answer 1

Correct Answer: 3

Given \( h = 4.25 \times 10^{-15} \, \text{eV} \cdot \text{s} \), the frequency of the incident light is related to the energy absorbed by the hydrogen atoms using the equation: \[ E = h \times f \] Where:
- \( E \) is the energy absorbed,
- \( h \) is Planck's constant,
- \( f \) is the frequency of the incident light.

The total emission lines are 6, so the electron must have absorbed energy and jumped from \( n = 1 \) to \( n = 4 \). The energy difference between these levels can be expressed as: \[ \Delta E = h \times f \] Using the formula for the frequency, we calculate: \[ \Delta E = 13.6 \left[ \frac{1}{4^2} - \frac{1}{2^2} \right] \, \text{eV} \] Now solving for the frequency: \[ f = \frac{12.75 \times 10^{-15}}{4.25 \times 10^{-15}} = 3 \times 10^{15} \, \text{Hz} \] Thus, \( x = 12 \).