frequency and kinetic energy

Question

the minimum frequency of light needed to eject electrons from a metal is called the threshold frequency.

a)Find the minimum energy needed to eject electrons from a metal with a threshold frequency of 4.87×10^14 s^−1. in J

b)With what maximum kinetic energy will electrons be ejected when this metal is exposed to light with a wavelength of 245 nm?

Answer 1

E = hν

E = energy = ?

h = Plancks constant = 6.626x10^-34 Js

ν = frequency = 4.87x1014s^-1

E = (6.626x10^-34 Js)(4.87x10¹⁴s^-1)

E = 3.23x10^-19 J

To find the maximum kinetic energy of the electrons ejected, we need to find the energy of the photons of wavelength 245 nm and then subtract this energy from the ionization energy determined in part (a) above. This is the amount of "extra" energy imparted to the ejected electron.

E = hν and c = λν so

E = hc / λ where c = speed of light (3x10⁸ m/s) and λ = wavelength in m = 245x10^-9 m

E = (6.626x10^-34 Js)(3x10⁸ m/s) / 245x10^-9 m

E = 8.11x10^-9 J

Some of this energy (3.23x10^-19 J) goes to ejecting the electron from the metal, and the remainder goes to the ejected electron as kinetic energy, KE.

So, KE of ejected electron = 8.11x10^-19 J - 3.23x10^-19 J = 4.88x10^-19 J