If the photon of the wavelength 150 pm strikes an atom and one of its inner bound electrons is ejected out with a velocity of 1.5 × 107 m s–1, calculate the energy with which it is bound to the nucleus.
Given:
Wavelength, λ = 150 pm
Velocity, v = 1.5 × 107 m s–1
Energy, E = hv = hc/λ
Where
n = number of photons emitted
h = Planck’s constant
c = velocity of radiation
λ = wavelength of radiation
Substituting the values in the given expression of Energy [E]:
= {6.626× 10-34] × [3× 108 ms-1]} / [150× 10-12m]
= [1.9878× 10-28] / [337.1× 10-9m]
= 1.3252 × 10-15 J
We know the formula for kinetic energy which is given as follows:
1/2 mv2 = kinetic Energy
Where
m = mass of electron
v = velocity of electron
K.E. = 1/2 × {[9.1× 10-31] × [1.5 × 107]2
= 1.02375× 10-16 J
Energy which bounded the electron to nucleus is:
= 13.25× 10-16J-1.025× 10-16J
=12.227× 10-16 J
= [12.227× 10-16]/ [1.602× 10-19]
=7.63× 103eV
Therefore, the energy which bounded the electron to nucleus is 7.63× 103eV.