Question

When monochromatic light of an unknown wavelength falls on a sample of silver, a minimum potential of 2.50 V is required to stop all of the ejected photoelectrons. Determine the (a) maximum kinetic energy and (b) maximum speed of the ejected photoelectrons. (c) Determine the wavelength in nm of the incident light. (The work function for silver is 4.73 eV.)

Answer 1

work done by stopping potential W = V*q

work done = change in KE = KEf - KEi

V*q = Kf - Kmax

Kf = 0

q = 1.6*10^-19 C

Kmax = V*q

Kmax = 2.5*1.6*10^-19 = 4*10^-19 J

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part(b)

kinetic energy K = (1/2)*m*v^2

4*10^-19 = (1/2)*9.11*10^-31*v^2

v = 9.37*10^5 m/s

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part(c)

from photo electric equation

Eincident = Wo + KEmax

hc/lambda = Wo + KEmax

KEmax = 4*10^-19 J

1 eV = 1.6*10^-19 J

Wo = 4.73 eV = 4.73*1.6*10^-19 J

h = planck's constant = 6.626*10^-34 Js

c = speed of light = 3*10^8 m/s

6.626*10^-34*3*10^8/lambda = (4.73*1.6*10^-19) + 4*10^-19

wavelength lambda = 172 nm <<----ANSWER