Question

The work function for silver is 4.7 eV.

(a) Find the threshold frequency and wavelength for the photoelectric effect to occur when monochromatic electromagnetic radiation is incident on the surface of a sample of silver.
Hz
nm

(b) Find the maximum kinetic energy of the electrons if the wavelength of the incident light is 180 nm.
  eV

(c) Find the maximum kinetic energy of the electrons if the wavelength of the incident light is 240 nm.
eV

Answer 1

1) The photoelectric equation involves;

      Here hf = Φ+E_k

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

     f = the frequency of the incident light in hertz (Hz)

    Φ = the work function in joules (J)

E_k = the maximum kinetic energy of the emitted electrons in joules (J)

. f = (Φ+E_k)/h

     = (7.5302x10^-19 J)/(6.63x10^-34 J s)

     = 1.135x10¹⁵ Hz.

Wavelength λ = c/f = (3*10⁸ m/s) /(1.135*10¹⁵ Hz)

                     = 264.317 nm.

B)

The equation for the photoelectric effect is

EK=hf–ø, where

EK is the maximum kinetic energy of the photoelectron
h is Planck's constant
f is the frequency of the light
ø is the work function of the metal

ø = 7.5302x10^-19 J

hf=hc/ λ = ((6.63x10^-34 J s)x(3x10⁸ m/s))/(180 nm)

hf=1.105x10^-18 J.

Then

E_K = (1.105x10^-18 J)-( 7.5302x10^-19 J) = 3.296x10^-19 J.

C)

ø = 7.5302x10^-19 J

hf=hc/ λ = ((6.63x10^-34 J s)x(3x10⁸ m/s))/(240 nm)

hf=8.2875x10^-19 J.

Then

E_K = (8.2875x10^-19 J)-( 7.5302x10^-19 J = 7.573x10^-20 J.