The work function for platinum is 6.35 eV. (a) Convert the value of the work fun
ID: 1898398 • Letter: T
Question
The work function for platinum is 6.35 eV.(a) Convert the value of the work function from electron volts to joules.
J
(b) Find the cutoff frequency for platinum.
Hz
(c) What maximum wavelength of light incident on platinum releases photoelectrons from the platinum's surface?
m
(d) If light of energy 7.2 eV is incident on platinum, what is the maximum kinetic energy of the ejected photoelectrons? Give the answer in electron volts.
eV
(e) For photons of energy 7.2 eV, what stopping potential would be required to arrest the current of photoelectrons?
V
Explanation / Answer
here,
work function for platinum , W = 6.35 eV
(a)
W = 6.35 eV
W = 6.35 * 1.6 * 10^-19 J
W = 1.02 * 10^-18 J
the work function in joules is 1.02 * 10^-18 J
(b)
the cut off frequency , f = W/h
f = 1.02 * 10^-18 /( 6.626 * 10^-34)
f = 1.54 * 10^15 Hz
the cutoff frequency is 1.54 * 10^15 Hz
(c)
the maximum wavelength of the photons , lamda = c/f
lamda = 3 * 10^8 /( 1.54 * 10^15)
lamda = 1.95 * 10^-7 m
the maximum wavelength of the incident light is 1.95 * 10^-7 m
(d)
energy of the incident light , E = 7.4 eV
the kinetic energy of the ejected photon KE = energy of inciident photons - work function
KE = 7.4 - 6.35
KE = 1.05 eV
teh kinetic energy of the ejected photelectrons is 1.05 eV
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