Question

When a metal was exposed to one photon of light at a frequency of 4.55× 1015 s–1, one electron was emitted with a kinetic energy of 4.10× 10–19 J. Calculate the work function of this metal. What is the maximum number of electrons that could be ejected from this metal by a burst of photons (at some other frequency) with a total energy of 5.11× 10–7 J?

Answer 1

As we know

h = 6.63 X 10^-34 Js ( Planck 's constant)

V

Given K.E = 4.10 X 10^-19J , Frequency = 4.55 X 10^15 s^-1

work Fuction = 6.63 X 10^-34 X 4.55 X 10^15s^-1 - (4.10 X10^-19 J) = 2.6 X 10^-18 J

b) Given Total Energy = 5.11 X 10^-7 J

Energy per electron = E = hf = 6.63 X 10^-34 X 4.55 X 10^15 = 30.16 X 10^-19

To find the maximum no. of electron we need binding energy

B.E = Energy per electron - K.E = 2.6 X 10^-18

maximum number of electrons = total eenergy / B.E = 5.11 X 10^-7J/ 2.6X10^-18 = 1.9 X 10^13 J