The work function for indium is 4.12 eV. (a) Convert the value of the work funct

Question

The work function for indium is 4.12 eV. (a) Convert the value of the work function from electron volts tojoules.
J

(b) Find the cutoff frequency for indium.
Hz

(c) What maximum wavelength of light incident on indium releases photoelectrons fromthe indium's surface?
m

(d) If light of energy 8.3 eV is incident on indium, what is the maximum kinetic energy of theejected photoelectrons? Give the answer in electron volts.
eV

(e) For photons of energy 8.3 eV, what stopping potential would berequired to arrest the current of photoelectrons?
V (a) Convert the value of the work function from electron volts tojoules.
J

(b) Find the cutoff frequency for indium.
Hz

(c) What maximum wavelength of light incident on indium releases photoelectrons fromthe indium's surface?
m

(d) If light of energy 8.3 eV is incident on indium, what is the maximum kinetic energy of theejected photoelectrons? Give the answer in electron volts.
eV

(e) For photons of energy 8.3 eV, what stopping potential would berequired to arrest the current of photoelectrons?
V

Explanation / Answer

   a.   1   eV   =   1.6* 10-19   J       Workfunction      =   4.12eV   =   4.12 * 1.6 *10-19          =   6.592* 10-19      J    b.   Cutofffrequency   f0   =   / h    f0   =   6.592* 10-19 / 6.62 *10-34   =   9.958 *1014   Hz    c.   Maximumwavelength   0   =   c/ f0   =   3.0 *108 / 9.958 * 1014          0   =   3.0126* 10-7   m    d.   Photoelectric equationis          Kmax   =   h*f   -      =   E   -       Here   E   =   Energyof incident photons    Kmax   =   8.3   -   4.12   =   4.18   eV    d.   Kmax   =   e* V0          Stoppingpotential   V0   =   4.18eV /e   =   4.18   V          Stoppingpotential   V0   =   4.18eV /e   =   4.18   V

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