An atom emits a photon of wavelength 1.08 meters. What is the energy change occurring in the atom due to this emission? (Planck's constant is 6.626 × 10È³Ê joule seconds, the speed of light is
2.998 × 10§ m/s)
the energy of the photon is hc/lamda where h is the Planck's constant, c is the speed of light and lamda is the wavelength in meters.
To proceed with the problem we need the formula:
E = hc/λ where E is the energy emitted by the photon (required), h is the Planck's constant (6.626 × 10-34 joule seconds), c is speed of light (2.998 × 108 m/s), and λ is the wave length (1.08 m). Plugging in the given to the equation and solve for energy, you would get 1.83*10^-25 J
1.8393*10-25joules
To calculate the energy change occurring in the atom due to the emission of a photon, you can use the equation:
Energy = h * (c / λ)
where:
- Energy is the energy change in the atom (in joules),
- h is Planck's constant (6.626 × 10^-34 joule seconds),
- c is the speed of light (2.998 × 10^8 m/s),
- λ (lambda) is the wavelength of the photon (in meters).
Substituting the given values, we have:
Energy = (6.626 × 10^-34 J s) * (2.998 × 10^8 m/s) / 1.08 m
Calculating this expression gives the energy change in joules.