Question

electrons are ejected from a metallic surface with speeds ranging from 4.5 times ten to the -5 when light with a wavelength of 600 nm is used a) what is the work function of the surface? B) what is the cutoff frequency for the surface?

Answer 1

When photons with Energy E strike the metallic surface we have this realation

E= Work function + Kinetic Enegy

E=   + KE

and energy E is

(where h is plank constant , c is speed of light and is wavelength)

E=3.31510^-19 J

and

Kinetic energy of ejected elcctron   

(where m_e is mass of electron and v is velocity of electron)

m_e= 9.1 x10^-31 Kg

V= 4.5x10^-5 m/s (unit not mentioned in problem is so taking m/s)

KE=92.13x10^-41 J

Now

= E - KE

=(3.31510^-19 ) -(92.13x10^-41)

=3.315x 10^-19 J   (approx)                       ( as the value of KE is very small)

(B)

cutoff frequency is the frequency at which electron ejects.

Energy requried by elctron to get ejected = work function ()

and we have

E =hf    ( where f is frequency )

here E=

= hf

f= /h

f=(3.315x 10^-19)/(6.63x10^-34)

f=510^-15 Hz (ANS)