The work function for platinum is6.35 eV. (a) Convert the value of the work func
ID: 1744279 • Letter: T
Question
The work function for platinum is6.35 eV. (a) Convert the value of the work function fromelectron volts to joules.1 J
(b) Find the cutoff frequency for platinum.
2 Hz
(c) What maximum wavelength of light incident on platinum releases photoelectrons from the platinum's surface?
3 m
(d) If light of energy 7.9 eV is incidenton platinum, what is the maximum kineticenergy of the ejected photoelectrons? Give the answer in electronvolts.
4 eV
(e) For photons of energy 7.9 eV, whatstopping potential would be required to arrest the current ofphotoelectrons?
5 V (a) Convert the value of the work function fromelectron volts to joules.
1 J
(b) Find the cutoff frequency for platinum.
2 Hz
(c) What maximum wavelength of light incident on platinum releases photoelectrons from the platinum's surface?
3 m
(d) If light of energy 7.9 eV is incidenton platinum, what is the maximum kineticenergy of the ejected photoelectrons? Give the answer in electronvolts.
4 eV
(e) For photons of energy 7.9 eV, whatstopping potential would be required to arrest the current ofphotoelectrons?
5 V
Explanation / Answer
(a) Convert the value of the work function fromelectron volts to joules. eV= V x qe ;qe=1.602x10-19 C or 1eV= 1.602x10-19 J W= 6.35 eV x1.602x10-19 C W=10.2x10-19 J(b) Find the cutoff frequency for platinum.
= hf0 f0=/h where h =4.135x10-15 eV s
f0=6.35/4.135x10-15 =
(c) What maximum wavelength of light incident on platinum releases photoelectrons from the platinum's surface?
= c/f0= 2.99x108/ f0=
(d) If light of energy 7.9 eV is incidenton platinum, what is the maximum kineticenergy of the ejected photoelectrons? Give the answer in electronvolts.
Ke= E - = (7.9 - 6.35) eV= 1.55 eV
(e) For photons of energy 7.9 eV, whatstopping potential would be required to arrest the current ofphotoelectrons? V= Ke = 1.55V
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