Question

A photon of wavelength 300 Å is incident upon a hydrogen atom at rest. The photon gives of its energy to the bound electron, thus releasing it from the atom (binding energy = 13.6 eV). Find the kinetic energy in eV and the velocity of the photoelectron.

Answer 1

Given parameters

wavelength $\lambda$ = 300 $\overset{0}{A}$

Binding energy = 13.6 eV = 2.178  $\times$ 10^-18 J

and we know that

speed of light C =   m/s

3 x 10

Now kinetic energy is given by

KE = Photon Energy - Binding energy

Photon energy =   

hc

h = Planck's Constant = 6.63 $\times$ 10^-34

$\lambda$ = 300 $\overset{0}{A}$ = 300 $\times$ 10^-10 meter

C =   m/s

3 x 10

Therefore Photon Energy =

hc

$= \frac{6.63\times 10^{-34}\times 3\times 10^{8}}{300\times 10^{-10} }$

Upon calculation we get photon energy = 6.63 $\times$ 10^-18

Therefore Kinetic Energy KE = Photon Energy - Binding energy

=  6.63 $\times$ 10^-18   -  2.178  $\times$ 10^-18

Kinetic Energy KE = 4.452 $\times$ 10^-18 J

Now

  $KE = \frac{1}{2}mv^{2}$

m = mass of hydrogen atom = 1.67 $\times$ 10^-27 Kg

Therefore

  $KE = \frac{1}{2}mv^{2}$

$4.452\times 10^{-18}= \frac{1}{2}\times 1.67\times 10^{-27}\times v^{2}$

  $v^{2}= \frac{4.452\times 10^{-18}}{8.35\times 10^{-28}}$

Upon calculation we get the velocity of the photoelectron v = 73 $\times$ 10³ Joule/ Kg

Therefore

Kinetic Energy KE = 4.452 $\times$ 10^-18 J

velocity of the photoelectron v = 73 $\times$ 10³ Joule/ Kg