A photon with a frequency of 5.02 × 1014 Hz is absorbed by an excited hydrogen a

Question

A photon with a frequency of 5.02 × 1014 Hz is absorbed by an excited hydrogen atom. This causes the electron to be ejected from the atom, forming an ion.
Calculate the energy of the photon in joules.
Determine the energy of the photon in electron volts.
What is the number of the lowest energy level (closest to the ground state) of a hydrogen atom that contains an electron that would be ejected by the absorption of this photon?
What is the kinetic energy of the ejected electron in electron volts?
What is the kinetic energy of the ejected electron in joules?
How fast is the electron moving?

Explanation / Answer

The energy of the photon in joules
E = hf     , where f is frequency = 5.02 × 1014 Hz , h = 6.625*10^-34

solving we get E = 33.25*10-20 J

energy in ev's = energy in joules /1.602*10-19

E = 2.075 ev

by absorbing this amont of energy an electron from n = 3 (E3 = -1.51ev )will be ejected

kinetic energy of ejected electron = energy of incident photons

K.E. = 2.075 ev

in joules K.E. = 33.25*10-20 J

speed of the electron

0.5 mv2 = 33.25*10-20 J

m = 9.1 * 10-31 kg

solving this

speed of the electron =8.54*105 m/s

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