Question

When light of frequency 4.96×10¹⁵ s^-1 shines on the surface of a metal, electrons are ejected with a maximum kinetic energy of 3.88×10^-19 J.

a) Calculate the wavelength of this light.

b) Find the binding energy of electrons to the metal.

c) What is the longest wavelength of light that will eject electrons from this metal?

Answer 1

a)To calculate the wavelength use the following equation:

(wavelength λ)*(frequency ν) = c (speed of light)
(wavelength λ)*( (4.96×10¹⁵ s^-1 )= 3.0 x 10^8 m/s.
wavelength = 6.05 x 10^-8 m

b)To calculate the binding energy of electrons to the metal use the following equation:

Ephoton = Ebinding + Ekinetic

The energy of the photon can be determined through the frequency:
Ephoton = h ν, where h is Planck's constant and ν is the frequency
Ephoton = (6.626 x 10^-34 Js)(4.96 x 10^15 1/s)
Ephoton = 3.286x 10^-18J

Ebinding = Ephoton - Ekinetic
Ebinding = 3.286 x 10^-18 J -3.88 x 10^-19 J
Ebinding = 2.898 x 10^-18J

c)

hv= work function; phi + KE

h = the Plank constant 6.63 x 10^-34 J s

v = the frequency of the incident light in hertz (Hz); 4.96×10¹⁵ s^-1

&phi = the work function in joules (J)

E_k = the maximum kinetic energy of the emitted electrons in joules (J) 3.88×10^-19 J.

hv= work function + KE

work function = 6.63 x 10^-34 J s*4.96×10¹⁵ s^-1 -3.88×10^-19 J.

work function =3.286x 10^-18J --3.88×10^-19 J = 2.898 J

h ν = work function + KE
For longest wavelength (or threshold frequency) then there will only be just enough energy for the photoelectrons to be liberated, hence their KE will be 0.
So h ν = work function

hv= 2.898 J

6.63 x 10^-34 J s v = 2.898 J

v= 4.37* x 10³³ /s

wavelength λ)*(frequency ν) = c (speed of light)
(wavelength λ)*( (4.37×10³³ s^-1 )= 3.0 x 10^8 m/s.
wavelength = 6.86 x 10^-26 m