Light Strikes Metal...Photoelectrons (PLEASE HELP! final <span title="tomorrow--desperate=understatement)?">tomorrow--desperate=under...</span>

Question

Light of frequency of 3.5 X 10^15 Hz strikes a metal and emits photoelectrons that have a max. energy of 5.9 eV. Calculate work function of the metal.

This class has no lecture and the book's a joke...any advice would be greatly greatly appreciated

Answer 1

You see, the electrons in an atom is attracted to the positive nucleus. So energy is required to pull out the electron. Light is an electromagnetic wave that possess enegy. The energy of a photon light of  frequency, f is given by

E = hf,

where h is a constant called Planck&#039;s constant

h= 6.626*10^-34 Js

E =  6.626*10^-34 Js * 3.5 X 10^15 Hz = 2.32*10^-18 J

This is the eenrgy of a photon of light.

Thes energy of one photon is absorbed by one electron. A part of this energy is used by the electron to overcome the attractive force of the nucleus(known as the work function of the metal). Once it is out of the nucleus, the remaining energy is used to impart kinetic energy to the electron(now known as the photoelectron)

So, the work function of the metal = Energy of the photon - Kinetic energy imparted to the electron.

The kinetic energy of the electron = 5.9 eV

= 5.9 * 1.6*10^-19  = 9.44*10^-19  J        (1 eV= 1.6*10^-19  J)

the work function of the metal = 2.32*10^-18 J - 9.44*10^-19  J

= 1.4*10^-18 J

= 1.4*10^-18 J/ 1.6*10^-19  

=8.6 eV

Answer 2

Work function of a metal   W = h .v

h = 6.625 x 10^ -34

Threhold frequency  v = 3.5 x 10^15 Hz

Therefore  W =  6.625 x 10^-34  x 3.5 x 10^15

                       = 1.98 x 10^-19 Joules

Maximum Kinetic energy K.E. =

           5.9 x 1.6 x 10^-19 J --  W