The work function for nickel is 5.15 eV. (a) Convert the value of the work funct
ID: 2299861 • Letter: T
Question
The work function for nickel is 5.15 eV.
(a) Convert the value of the work function from electron volts to joules.
J
(b) Find the cutoff frequency for nickel.
Hz
(c) What maximum wavelength of light incident on nickel releases photoelectrons from the nickel's surface?
m
(d) If light of energy 7.7 eV is incident on nickel, what is the maximum kinetic energy of the ejected photoelectrons?
Give the answer in electron volts.
eV
(e) For photons of energy 7.7 eV, what stopping potential would be required to arrest the current of photoelectrons?
V
Explanation / Answer
a) 1 eV = 1.6 * 10E-19 C * 1 V = 1.6 * 10E-19 J
5.15 * 1.6 * 10E-19 = 8.24 * 10E-19 J
b) E = h f - E0 where E0 is the work function
E = 0 or h * f0 = E0 KE of ejected electrons = 0
f0 = E0 / h = 8.24 * 10E-19 / 6.63 * 10E-34 = 1.24 * 10E15
c) wavelength = c / f 0 = 3 * 10E8 / 1.24 * 10E15 = 2.41 * 10E-7 m
d) E = h * f - E0 = 7.7 eV - 5.15 eV = 2.55 eV maximum KE
e) E = 2.55 eV required stopping potential
2.55 eV * 1.6 * 10E-19 J / eV = 4.08 * 10E-19 J stopping potential in Joules
4.08 * 10E-19 J / 1.6 * 10E-19 C = 2.55 J / C = 2.55 V
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