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 released during an energy level transition is calculated using \( E = h u \). Here, \( E \) represents the energy of the emitted radiation, given as \( 2.55 \, \text{eV} \). \( h \), Planck's constant, is \( 6.626 \times 10^{-34} \, \text{J} \cdot \text{s} \), and \( u \) is the frequency of the emitted radiation. Frequency \( u \) is connected to wavelength \( \lambda \) by \( u = \frac{c}{\lambda} \), where \( c \) is the speed of light (\( 3.0 \times 10^8 \, \text{m/s} \)) and \( \lambda \) is the wavelength. Substituting \( u \) into the energy equation yields \( E = h \frac{c}{\lambda} \). Rearranging to solve for \( \lambda \) gives \( \lambda = \frac{h c}{E} \). Using the given values: \( h = 6.626 \times 10^{-34} \, \text{J} \cdot \text{s} \), \( c = 3.0 \times 10^8 \, \text{m/s} \), and \( E = 2.55 \, \text{eV} = 2.55 \times 1.602 \times 10^{-19} \, \text{J} \). The calculation for wavelength is \( \lambda = \frac{6.626 \times 10^{-34} \times 3.0 \times 10^8}{2.55 \times 1.602 \times 10^{-19}} \), which results in \( \lambda = \frac{1.9878 \times 10^{-25}}{4.0881 \times 10^{-19}} = 4.87 \times 10^{-7} \, \text{m} \). Consequently, the emitted radiation has a wavelength of \( \lambda = 487 \, \text{nm} \).