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)
7.1560*10-34
To find 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, c is the speed of light, and λ (lambda) is the wavelength of the photon.
Given:
Planck's constant (h) = 6.626 × 10^-34 joule seconds
Speed of light (c) = 2.998 × 10^8 m/s
Wavelength (λ) = 1.08 meters
Now, let's substitute these values into the equation:
E = (6.626 × 10^-34 joule seconds) x (2.998 × 10^8 m/s) / (1.08 meters)
First, let's simplify the equation:
E = (6.626 x 10^-34 x 2.998 x 10^8) joule seconds / 1.08 meters
Next, let's perform the multiplication:
E = 19.82 x 10^-26 joule seconds / 1.08 meters
Now, let's divide the numerator by the denominator:
E = 1.83 x 10^-25 joules
Therefore, the energy change occurring in the atom due to the emission is approximately 1.83 x 10^-25 joules.