Question

A beam of ultraviolet radiation, with frequency of 2.570×1015 Hz is directed on a clean metal surface. The work function of the metal is 3.40 eV. Calculate the maximum kinetic energy for an electron dislodged from the surface by the radiation.

Answer 1

frequency f = 2.57 x 10¹⁵ Hz ,

work function = 3.40 eV

energy E = hf = 6.625 x 10^-34 J-s x 2.57 x 10¹⁵ = 17.026 x 10^-19 J = 10.63 eV

from einstiens relation

E = work function + kinetic energy

kinetic energy = 10.63 eV- 3.40eV = 7.23 eV = 11.58 x 10^-19 J

Answer 2

frequency f = 2.57 x 1015 Hz ,
work function = 3.40 eV
energy E = hf = 6.625 x 10-34 J-s x 2.57 x 1015 = 17.026 x 10-19 J = 10.63 eV
from einstiens relation
E = work function + kinetic energy
kinetic energy = 10.63 eV- 3.40eV = 7.23 eV = 11.58 x 10-19 J

Answer 3

photoelectric equation
hv = phi + 0.5 m v(max)^2
photon energy = work function + max KE
KE (max) = hv - phi
KE (max) = 6.6*10^-34*2.15*10^15 - 3.40*1.6*10^-19 (Joule)
KE (max) = 14.19*10^-19 - 5.44*10^-19
KE (max) = 8.75*10^-19 Joule