The work function for platinum is 6.35 eV. (a) Convert the value of the work fun
ID: 1744777 • Letter: T
Question
The work function for platinum is 6.35 eV. (a) Convert the value of the work function from electron volts tojoules.J
(b) Find the cutoff frequency for platinum.
Hz
(c) What maximum wavelength of light incident on platinum releases photoelectrons fromthe platinum's surface?
m
(d) If light of energy 7.2 eV isincident on platinum, what is themaximum kinetic energy of the ejected photoelectrons? Give theanswer in electron volts.
eV
(e) For photons of energy 7.2 eV, what stopping potential would be requiredto arrest the current of photoelectrons?
V Please Help (a) Convert the value of the work function from electron volts tojoules.
J
(b) Find the cutoff frequency for platinum.
Hz
(c) What maximum wavelength of light incident on platinum releases photoelectrons fromthe platinum's surface?
m
(d) If light of energy 7.2 eV isincident on platinum, what is themaximum kinetic energy of the ejected photoelectrons? Give theanswer in electron volts.
eV
(e) For photons of energy 7.2 eV, what stopping potential would be requiredto arrest the current of photoelectrons?
V Please Help
Explanation / Answer
a. 1 eV = 1.6* 10-19 J Workfunction = 6.35 eV = 6.35* 1.6 * 10-19 = 1.016* 10-18 J b. = h* f0 cut offfrequency f0 = 10.16* 10-19 / 6.62 *10-34 = 1.534 *1015 Hz c. Maximum (threshold)wavelength 0 = c/ f0 = 3.0 *108 / 1.534 * 1015 0 = 1.956* 10-7 m d. Photoelectric equationis Kmax = h* f - max. k.e. ofphotoelectrons Kmax = 7.2eV - 6.35 eV = 0.85 eV e. Also Kmax = e* V0 0.85eV = e * V0 Stoppingpotential V0 = 0.85 V Stoppingpotential V0 = 0.85 VRelated Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.