An electron falls from an energy level of -5.44eV to another energy level, E. If the emitted photon is of wavelength 5.68 x 10−6 − 6 m, calculate the energy...

An electron falls from an energy level of -5.44eV to another energy level, E. If the emitted photon is of wavelength 5.68 x 10−6 ${}^{-6}$m, calculate the energy change. [ Plank's constant = 6.63 x 10−34 ${}^{-34}$Js, emitted radiation speed = 3.0 x 108 ${}^8$ms−1 ${}^{-1}$]

Answer Details

To find the energy change when an electron falls from one energy level to another, we need to calculate the energy of the emitted photon. This energy can be found using the formula:

E = hν   or   E = hc/λ

where:

• h is Planck's constant, 6.63 x 10^-34 Js
• c is the speed of light, 3.0 x 10⁸ ms^-1
• λ is the wavelength of the emitted photon, 5.68 x 10^-6 m

Substitute these values into the equation:

E = (6.63 x 10^-34 Js) * (3.0 x 10⁸ ms^-1) / (5.68 x 10^-6 m)

First, calculate the numerator:

(6.63 x 10^-34) * (3.0 x 10⁸) = 1.989 x 10^-25 J·m

Then, divide by the wavelength:

E = 1.989 x 10^-25 J·m / 5.68 x 10^-6 m = 3.5 x 10^-20 J

Therefore, the energy change when the electron falls is approximately 3.5 x 10^-20 J.

Checking the options provided, the closest value is 3.49 x 10^-20 J.