Question

The work function for lithium is 300 kJ/mole. What is the wavelength of light that will remove an electron from an atom in lithium metal so that it has an ejected energy of 0.5 eV once free of the lithium atom?

Answer 1

Ans:

Given, work function = 300 kJ/mol ==> 49.809 x 10^-20 J/atom of lithium (0.3/avagadro's number)

Kinetic energy of ejected electron (ejected energy) = 0.5 eV = 8.01 x 10^-20 J

Energy (E) associated with the light of wavelength (λ); E = hc/ λ

{where, h = planks constant = 6.626 x 10^-34 J s & c = speed of light = 3 x 10⁸ m s-¹ }

According to photo electric effect, Energy (E) associated with the light of wavelength (λ) will be given by,

E = Work function + Kinetic energy of ejected electron (ejected energy)

hc/ λ = (49.809 x 10^-20 J) + (8.01 x 10^-20 J)

Substituting for h = 6.626 x 10^-34 J s & c = 3 x 10⁸ m s^-1, one will get

λ = 3.4379 x 10^-7 m

Wavelength of the light will be = λ = 3.4379 x 10^-7 m