Question

A metal surface is susceptible to a photoelectric threshold frequency of 8.529*10>14Hz. Will the monochromatic UV light sources of wavelengths 283 nm, 342 nm, and 363 nm be able to eject photoelectrons from the metal. If so what is the speed of the fastest electron ejected?

Answer 1

for the monochromatic lights to be able to eject electrons they must have the minimum threshold energy
i.e.
E(threshold) = hf = 6.626e-34 * 8.529e14
E(threshold) = 56.51e-20 J

E(283) = hc/lambda = 6.626e-34 * 3e8 / 283e-9 = 70.24e-20 J
E(342) = hc/lambda = 6.626e-34 * 3e8 / 342e-9 = 58.12e-20 J
E(363) = hc/lambda = 6.626e-34 * 3e8 / 363e-9 = 54.76e-20 J
as can be seen from the results

E(283) and E(342) are greater than E(threshold) hence they will be able to remove electron from the metal surface

Now for fastest electron to emit, it must have highest electron energy which is nothing but gained due to the monochromatic light hitting the elctron (some enrgy is lost called threshold energy and rest is imparted to emitted electron)

i.e
E(electron) = E(283) - E(thershold)
E(electron) = 70.24e-20 - 56.51e-20
E(electron) = 13.73e-20

now
E(electron) = 1/2mv^2
v^2 = 2*E(electron)/m
v^2 = 2*13.73e-20 / 9.1e-31
v^2 = 30.18e10
v = 5.49e5 m/s

E(electron) = E(342) - E(thershold)
E(electron) = 58.12e-20 - 56.51e-20
E(electron) = 1.61e-20

now
E(electron) = 1/2mv^2
v^2 = 2*E(electron)/m
v^2 = 2*1.61/2/9e-20 / 9.1e-31
v^2 = 3.54e10
v = 1.88e5 m/s