Calculate the change in energy of an atom that emits a photon of wavelength 2.21 meters. (Planck’s constant is 6.626 x 10-34 joule seconds, the speed of light is 2.998 x 108 m/s)
h f
where h = 6.626 * 10^-34
f = 1/T = c/lambda = 2.998*10^8/2.21
so
E = 6.626*10^-34 * 2.998*10^8/2.21 Joules
dont
To calculate the change in energy of an atom that emits a photon of a specific wavelength, we can use the equation:
ΔE = hc/λ
Where:
ΔE is the change in energy
h is Planck's constant (6.626 x 10^-34 joule seconds)
c is the speed of light (2.998 x 10^8 m/s)
λ is the wavelength of the photon
Now, let's substitute in the given values and solve the equation:
ΔE = (6.626 x 10^-34 joule seconds) * (2.998 x 10^8 m/s) / (2.21 meters)
Performing the calculation, we have:
ΔE = 8.94 x 10^-19 joules.
Therefore, the change in energy of the atom that emits a photon with a wavelength of 2.21 meters is approximately 8.94 x 10^-19 joules.