The wavelength of the light emitted by a hydrogen atom during a transition from \( n = 3 \) to \( n = 2 \) is \( 656.3 \, \text{nm} \). What is the energy of the photon emitted during this transition?
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When calculating the energy of a photon, always use the correct units for wavelength (meters), Planck's constant, and the speed of light. The wavelength in nanometers should be converted to meters.
To calculate the energy of the photon emitted, we use the formula that relates the energy of a photon to its wavelength:
\[
E = \frac{hc}{\lambda}
\]
Where:
- \( E \) is the energy of the photon,
- \( h = 6.626 \times 10^{-34} \, \text{J·s} \) is Planck's constant,
- \( c = 3.0 \times 10^8 \, \text{m/s} \) is the speed of light,
- \( \lambda = 656.3 \times 10^{-9} \, \text{m} \) is the wavelength of the light.
Substituting the known values:
\[
E = \frac{6.626 \times 10^{-34} \times 3.0 \times 10^8}{656.3 \times 10^{-9}}
\]
\[
E = \frac{1.9878 \times 10^{-25}}{656.3 \times 10^{-9}} = 3.02 \times 10^{-19} \, \text{J}
\]
Thus, the energy of the photon emitted during the transition is \( 3.02 \times 10^{-19} \, \text{J} \).
