The work function for platinum is 6.35 eV. (a) Convert the value of the work fun
ID: 1949746 • 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 9 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 9 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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