The work function for platinum is6.35 eV. (a) Convert the value of the work func
ID: 1743904 • Letter: T
Question
The work function for platinum is6.35 eV. (a) Convert the value of the work function fromelectron 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 8.1 eV is incidenton platinum, what is the maximum kineticenergy of the ejected photoelectrons? Give the answer in electronvolts.
eV
(e) For photons of energy 8.1 eV, whatstopping potential would be required to arrest the current ofphotoelectrons?
V (a) Convert the value of the work function fromelectron 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 8.1 eV is incidenton platinum, what is the maximum kineticenergy of the ejected photoelectrons? Give the answer in electronvolts.
eV
(e) For photons of energy 8.1 eV, whatstopping potential would be required to arrest the current ofphotoelectrons?
V
Explanation / Answer
6.35ev=6.35*1.6*10-19J=10.16*10-19J 10.16*10-19 isthe cut off energy of the radiation=h so cut off frequency=1.533*1015Hz maximum wavelength corresponds to minimum wavelength=c/ c=speed of light=3*108m/s/1.533*1015Hz=1.956*10-7 the excess energy suplied above work functionwill be converted in to K.Eso K.E=8.1-6.35=1.75e.v to stop the current Vq=K.E so 2.8 volts
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