Question

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)

Answer 1

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.

Answer 2

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

Answer 3

1.8393*10-25joules