Question

Light of wavelength 400 nm is shone on a metal surface in an apparatus like that of fig. 2.9. The work function of the metal is 2.50 ev. (a) find the extinction voltage, that is, the retarding voltage at which the photoelectron current disappears. (b) find the speed of the fastest photoelectrons

Answer 1

a) the retarding voltage at which the photoelectron current disappears:

• E= 12400/λ  =  12400/4000 A°

     = 3.1 eV

b) the speed of the fastest photoelectrons:

• E=W₀ + K.E

    3.1=2.50+ 1/2mv^2

    1/2mv^2=0.6eV

                = 0.6 * 1.6*10^-19

                =9.6*10^-20

             v^2= (9.6*10^-20 * 2)/m

 where, m= mass of electron = 9.1*10^-31

 v=((9.6*10^-20 * 2)/ 9.1*10^-31)^1/2

      = 4.58*10^-5m/s