Question

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?

• A. \( 3.02 \times 10^{-19} \, \text{J} \)
• B. \( 4.56 \times 10^{-19} \, \text{J} \)
• C. \( 2.18 \times 10^{-19} \, \text{J} \)
• D. \( 5.00 \times 10^{-19} \, \text{J} \)

Hint:

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.

Answer 1

The Correct Option is A

The photon's energy is computed using the formula \( E = \frac{hc}{\lambda} \). In this equation, \( E \) represents the photon energy, \( h = 6.626 \times 10^{-34} \, \text{J·s} \) is Planck's constant, \( c = 3.0 \times 10^8 \, \text{m/s} \) is the speed of light, and \( \lambda = 656.3 \times 10^{-9} \, \text{m} \) is the light's wavelength. Substituting these values yields: \[ E = \frac{6.626 \times 10^{-34} \times 3.0 \times 10^8}{656.3 \times 10^{-9}} \] \[ E = \frac{1.9878 \times 10^{-25}}{656.3 \times 10^{-9}} = 3.02 \times 10^{-19} \, \text{J} \] Therefore, the emitted photon's energy is \( 3.02 \times 10^{-19} \, \text{J} \).