Question

When a metal was exposed to photons at a frequency of 3.51× 1015 s–1, electrons were emitted with a maximum kinetic energy of 5.60× 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 1.78× 10–7 J?

Answer 1

1)

lets find the energy of photon

use:

E = h*f

=(6.626*10^-34 J.s)*(3.51*10^15 s-1

= 2.326*10^-18 J

KE = 5.60*10^-19 J

use:

KE = E - Eo

5.60*10^-19 = 2.326*10^-18 - Eo

Eo = 1.77*10^-18 J

this i work function

Answer: 1.77*10^-18 J

2)

each photon can release 1 electron

So, maximum number of electrons emitted = maximum number of photons

Lets calculate the number of photons in 1.78*10^-7 J

energy of 1 photon = 2.326*10^-18 J

number of photons = total energy / energy of 1 photon

= (1.78*10^-7)/(2.326*10^-18)

= 7.65*10^10

This is also number of electrons emitte

Answer: 7.65*10^10