EBK COLLEGE PHYSICS
EBK COLLEGE PHYSICS
11th Edition
ISBN: 8220103600385
Author: Vuille
Publisher: Cengage Learning US
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Chapter 27, Problem 11P

(a)

To determine

The work function in Joules.

(a)

Expert Solution
Check Mark

Answer to Problem 11P

The work function in Joules is 1.02×1018J .

Explanation of Solution

The work function in joules is,

ϕJ=ϕeV(1.60×1019J1eV)

    • ϕJ is the work function in joules
    • ϕeV is the work function in electron volt

Substitute 6.35eV for ϕeV .

ϕJ=(6.35eV)(1.60×1019J1eV)=1.02×1018J

Conclusion:

Thus, the work function in Joules is 1.02×1018J .

(b)

To determine

The cutoff frequency for platinum.

(b)

Expert Solution
Check Mark

Answer to Problem 11P

The cutoff frequency for platinum is 1.53×1015Hz .

Explanation of Solution

The energy of the photon is,

Ephoton=hfc

    • h is the Plank’s constant
    • fc is the cutoff frequency

The cutoff frequency is,

fc=ϕh

Substitute 6.35eV for ϕ and 6.63×1034J.s for h .

fc=6.35eV6.63×1034J.s(1eV1.60×1019J)=1.53×1015Hz

Conclusion:

Thus, the cutoff frequency for platinum is 1.53×1015Hz .

(c)

To determine

The maximum wavelength of the light incident on the platinum surface.

(c)

Expert Solution
Check Mark

Answer to Problem 11P

The maximum wavelength of the light incident on the platinum surface is 196nm .

Explanation of Solution

The equation for cutoff wavelength is,

λc=hcϕ=cfc

Substitute 3.00×108m/s for c and 1.53×1015Hz for fc .

λc=3.00×108m/s1.53×1015Hz=1.96×107m=196nm

Conclusion:

Thus, the maximum wavelength of the light incident on the platinum surface is 196nm .

(d)

To determine

The maximum kinetic energy of the ejected photoelectrons.

(d)

Expert Solution
Check Mark

Answer to Problem 11P

The maximum kinetic energy of the ejected photoelectrons is 2.15eV .

Explanation of Solution

The equation for maximum kinetic energy is,

KEmax=Ephotonϕ

    • Ephoton is the energy of the photon

Substitute 8.50eV for Ephoton and 6.35eV for ϕ .

KEmax=8.50eV6.35eV=2.15eV

Conclusion:

Thus, the maximum kinetic energy of the ejected photoelectrons is 2.15eV .

(e)

To determine

The stopping potential required to arrest the current of photoelectrons.

(e)

Expert Solution
Check Mark

Answer to Problem 11P

The stopping potential required to arrest the current of photoelectrons is 2.15V .

Explanation of Solution

We have the relation,

eVs=KEmaxVs=KEmaxe

Substitute 2.15eV for KEmax and 1.60×1019C .

Vs=2.15eV1.60×1019C=2.15V

Conclusion:

Thus, the maximum kinetic energy of the ejected photoelectrons is 2.15eV

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