The stopping potential in a photoelectric experiment is 1.8V,?

Question

when the illuminating radiation has wavelength 365nm.

a) What is the work function of the emitting surface?

b) What would be the stopping potential for 280nm radiation?

Show all working please >.<

Answer 1

i hope this is correct.

(we just recently taken up this topic in our school)

a.

Kmax=e*StoppingPotential

          =(-1.6x10^-19)(1.8) = -2.88x10^-19 J

f = c/wavelegnth = 8.22x10^14 Hz

WorkFunc= -Kmax + hf

                  = 2.88x10^-19 + 6.62610^-34(8.22x10^14)

                  = 8.33x10^-19 J

                  = 5.2 eV

final answer: 5.2 eV

b.

still use the formulae:

f = c/wavelength

Kmax = hf - WorkFunc

StoppingPotential = Kmax/e