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

To solve the problem, we determine the energy transition in a hydrogen atom. Given that there are six different wavelengths emitted, this implies transitions between energy levels involve \( n = 4 \) down to \( n = 1 \), with six transitions possible: \(4 \rightarrow 3\), \(4 \rightarrow 2\), \(4 \rightarrow 1\), \(3 \rightarrow 2\), \(3 \rightarrow 1\), and \(2 \rightarrow 1\). The highest transition, from \( n = 4 \) to \( n = 1 \), represents the possible initial absorption followed by eventual emission. The energy difference between levels is given by the formula: \[ E_{n} = -13.6\, \text{eV} \times \left(\frac{1}{n^2}\right) \] The initial absorption energy is: \[ \Delta E = E_1 - E_4 = -13.6\, \text{eV} \left(1 - \frac{1}{16}\right) = -13.6 \times \frac{15}{16} = -12.75 \, \text{eV} \] Convert this energy to frequency using: \[ E = h \nu \rightarrow \nu = \frac{12.75 \times 1.602 \times 10^{-19}\, \text{J}}{6.626 \times 10^{-34}\, \text{J s}} \] \[ \nu = 3.087 \times 10^{15} \, \text{Hz} \] Thus, \( x \approx 3.087 \), confirming \( x = 3,3 \), lies within the range. Therefore, \( x = 3.1 \) to match the value with precision.