Question

Two energy levels of an electron in a hydrogen atom are separated by 2.55 eV. Find the wavelength of radiation emitted when the electron makes a transition from the higher energy level to the lower energy level.

Hint:

The wavelength of radiation emitted during an electronic transition can be calculated using the energy-wavelength relation. Don't forget to convert the energy from eV to joules before using the formula.

Answer 1

The energy of a photon emitted during a transition between two energy levels is given by the equation: \[ E = h \nu \] where:
\( E \) is the energy of the emitted radiation (in this case, \( E = 2.55 \, \text{eV} \)),
\( h = 6.626 \times 10^{-34} \, \text{J} \cdot \text{s} \) is Planck's constant,
\( \nu \) is the frequency of the emitted radiation.
The frequency \( \nu \) is related to the wavelength \( \lambda \) by the equation:
\[ \nu = \frac{c}{\lambda} \] where:
\( c = 3.0 \times 10^8 \, \text{m/s} \) is the speed of light,
\( \lambda \) is the wavelength of the emitted radiation.
Substitute \( \nu \) into the energy equation:
\[ E = h \frac{c}{\lambda} \] Rearranging for \( \lambda \):
\[ \lambda = \frac{h c}{E} \] Now, substitute the given values:
\( h = 6.626 \times 10^{-34} \, \text{J} \cdot \text{s} \),
\( c = 3.0 \times 10^8 \, \text{m/s} \),
\( E = 2.55 \, \text{eV} = 2.55 \times 1.602 \times 10^{-19} \, \text{J} \).
Thus:
\[ \lambda = \frac{6.626 \times 10^{-34} \times 3.0 \times 10^8}{2.55 \times 1.602 \times 10^{-19}} \] \[ \lambda = \frac{1.9878 \times 10^{-25}}{4.0881 \times 10^{-19}} = 4.87 \times 10^{-7} \, \text{m} \] Therefore, the wavelength of the radiation emitted is \( \lambda = 487 \, \text{nm} \).