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 × 10-34 joule seconds, the speed of light is 2.998 × 108 m/s)
dE = hc/wavelength.
Plug and chug.
To calculate the energy change occurring in the atom due to the emission of a photon, we can use the equation:
E = hc/λ
where:
E is the energy change
h is Planck's constant (6.626 × 10^-34 J·s)
c is the speed of light (2.998 × 10^8 m/s)
λ is the wavelength of the photon (1.08 meters)
Substituting the given values into the equation, we have:
E = (6.626 × 10^-34 J·s * 2.998 × 10^8 m/s) / 1.08 meters
Simplifying the equation:
E = (1.989 × 10^-25 J·m) / 1.08 meters
E = 1.843 × 10^-25 J
Therefore, the energy change occurring in the atom due to this emission is 1.843 × 10^-25 Joules.