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)
dE = hc/wavelength
To calculate the energy change occurring in the atom due to the emission of a photon, you can use the equation:
Energy (E) = Planck's constant (h) multiplied by the speed of light (c) divided by the wavelength (λ)
Given:
Planck's constant (h) = 6.626 × 10^-34 Joule seconds
Speed of light (c) = 2.998 × 10^8 meters per second
Wavelength (λ) = 1.08 meters
Let's plug in the values and solve for the energy change:
E = (h * c) / λ
E = (6.626 × 10^-34 Joule seconds * 2.998 × 10^8 meters per second) / 1.08 meters
First, multiply the values in the numerator:
E = 1.987348 × 10^-25 Joules meter
Then, divide that result by the value in the denominator:
E ≈ 1.84 × 10^-25 Joules
Therefore, the energy change occurring in the atom due to the emission of a photon with a wavelength of 1.08 meters is approximately 1.84 × 10^-25 Joules.