Question

When a metal was exposed to photons at a frequency of 4.90× 1015 s–1, electrons were emitted with a maximum kinetic energy of 4.00× 10–19 J. Calculate the work function of this metal. 2. 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 6.89× 10–7 J?​

Answer 1

1)

we have:

frequency = 4.9*10^15 s-1

we have below equation to be used:

Energy = Planck constant*frequency

=(6.626*10^-34 J.s)*(4.9*10^15) s-1

= 3.25*10^-18 J

Now use:

E = Eo + KE

Eo = E - KE

= (3.25*10^-18 J) - (4.00*10^-19 J)

= 2.85*10^-18 J

Answer: 2.85*10^-18 J

2)

maximum number of photons = Energy / Eo

= (6.89*10^-7)/(2.85*10^-18)

= 2.42*10^11

1 photon can release 1 electron

Answer: 2.42*10^11