Question

Hydrogen atom in its ground energy state absorbs a photon, which excites it to an energy level of \(n = 4\). Calculate the frequency of photon.

Hint:

The energy of a photon can be calculated from the energy difference between two energy levels in an atom. Use the equation \(E = h \nu\) to find the frequency of the photon.

Answer 1

The energy levels of the hydrogen atom are quantized, and the energy corresponding to a particular energy level \(n\) is given by the following formula derived from the Bohr model:
\[ E_n = - \frac{13.6 \, \text{eV}}{n^2} \] Where: - \(E_n\) is the energy of the \(n\)-th energy level,
- \(n\) is the principal quantum number,
- \(13.6 \, \text{eV}\) is the Rydberg energy constant for the hydrogen atom.
The energy difference \(\Delta E\) between the ground state (\(n = 1\)) and the excited state (\(n = 4\)) is given by:
\[ \Delta E = E_4 - E_1 = \left( - \frac{13.6}{4^2} \right) - \left( - \frac{13.6}{1^2} \right) \] Simplifying:
\[ \Delta E = - \frac{13.6}{16} + 13.6 = 13.6 \left( 1 - \frac{1}{16} \right) \] \[ \Delta E = 13.6 \times \frac{15}{16} = 12.75 \, \text{eV} \] The energy of the photon absorbed by the atom is equal to the energy difference, so the energy of the photon is \(12.75 \, \text{eV}\).
To calculate the frequency \(\nu\) of the photon, we use the relationship between energy and frequency:
\[ E = h \nu \] Where: - \(E\) is the energy of the photon,
- \(h\) is Planck's constant (\(h = 6.626 \times 10^{-34} \, \text{J} . \text{s}\)),
- \(\nu\) is the frequency of the photon.
We first convert the energy from eV to joules. Since \(1 \, \text{eV} = 1.602 \times 10^{-19} \, \text{J}\), we get:
\[ E = 12.75 \, \text{eV} = 12.75 \times 1.602 \times 10^{-19} \, \text{J} = 2.04 \times 10^{-18} \, \text{J} \] Now, solving for the frequency:
\[ \nu = \frac{E}{h} = \frac{2.04 \times 10^{-18}}{6.626 \times 10^{-34}} \, \text{Hz} \] \[ \nu \approx 3.08 \times 10^{15} \, \text{Hz} \] Thus, the frequency of the photon is approximately \(3.08 \times 10^{15} \, \text{Hz}\).