Question

^{In the photoelectric effect, you got familiar with threshold frequency, stopping potential, and how they are inter-related. The wavelength() corresponding to the threshold frequency of potassium is 558 nm. What is the work function for potassium? What is the maximum kinetic energy of e- when light of 400 nm is incident on potassium? If the intensity of light is 10-2W/m², how many photons are incident per second per square meter? (4 points)}

Answer 1

If wavelength of the incident light is and work function is Then maximum KE of the electron emitted KE max = hc - ¢ Here Work function $=m where

Ath is threshold wavelength If stopping potential is Vo then KE max = eV, ev=h hc threshold wavelength is given as 558 nm 4 = hc d 6.626 x 10-34 x 3 x 108 558 x 109

$= he d. = 6.626 x 10-34 x 3 x 108 -= 2.226 x eV (558 x 10°) (1.6 x 10-19) (1xev=1,6x 10-19) When 2 = 400nm KE = hc - 4 6.626 X 10-34 x 3 x 10°_.2226=0.88 x eV KE max = (400 10"")> (1.6 x 10-19) Intensity of light

1 = + = 102 A=lmº t=1 E = 10? At = 102 x (1)>(1)= 10-2 if no of photon incident are n then Eenmaa n= Ea hc ne (102)(400 x 10°) (6.62 x 10-34)> (3 x 109) -2.014 x 1016 n= 2.014 x 1016