The work function for platinum is 6.35eV. (a.) Convert value of the work functio
ID: 2014800 • Letter: T
Question
The work function for platinum is 6.35eV.(a.) Convert value of the work function from electron volts to joules.
(b.) Find the cutoff frequency for platinum.
(c.) What maximum wavelength of light incident on platinum releases photoelectrons from the platinum's surface?
(d.) If light of energy 8.50 eV is incident on zinc, what is the maximum kinetic energy of the ejected photoelectrons? Give the answer in electron volts.
(e.) For photons of energy 8.50 eV, what stopping potential would be required to arrest the current of photoelectrons?
Explanation / Answer
A) 6.35X 1.6x10^-19 = 3.81x10^-18 J B) E= hf f =E/h = 6.35/4.14X 10^-15 = 1.5338X10^15 Hz C) C= f lamda lamda= c/f = 3x10^8/ 1.53x10^15 = 1.96X10^-7m D) What is the work function for zinc ???? KE= hf- work function Stopping potential => E=qV V=E/q
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