Recitation 5Chapter 28

Problem 28.4. Calculate the energy, in electron volts, of a photon whose frequency is (a) 620 THz, (b) 3.10 GHz, and(c) 46.0 MHz. (d) Determine the corresponding wavelengths for these photons and state the classification of each on theelectromagnetic spectrum.

(a) (b) (c)E = hf 2.56 eV 12.8 µeV 190 neVλ = c/f 484 nm 9.67 cm 6.51 m

Class Visible (blue) Microwave Radio

See Figure 24.12 (page 823) in the text for a classification spectrum.

Problem 28.6. The average threshold of dark-adapted (scotopic) vision is 4.00 · 10−11 W/m2 at a central wavelength of500 nm. If light having this intensity and wavelength enters the eye and the pupil is open to its maximum diameter of8.50 mm, how many photons per second enter the eye?

The total power into the eye is

P = IA = πr2I =πd2I

4= 2.27 fW , (1)

and the energy per photon is

E = hf =hc

λ= 2.48 eV = 3.97 · 10−19 J , (2)

so the number of photons entering per second is

Φp =P

E= 5710 . (3)

Problem 28.9. Molybdenum has a work function of 4.20 eV. (a) Find the cutoff wavelength and cutoff frequency for thephotoelectic effect. (b) What is the stopping potential if the incident light has a wavelength of 180 nm?

(a) The cutoff wavelength is the wavelength where the incoming light has barely enough energy to free an electron, i.e. all ofthe photon’s energy goes into overcoming the work function barrier.

hf = φ (4)

f =φ

h=

4.20 eV · 1.60 · 10−19 J/eV6.63 · 10−34 J·s

= 1.01 PHz (5)

λf = c (6)

λ =c

f= 296 nm (7)

(b) The photon brings in hf , but much of that energy goes to overcoming the work function barrier. The left over energyhf−φ becomes the electron’s kinetic energy. The stopping potential is the voltage change which matches that kinetic energy.

Kmax = hf − φ = hc

λ− φ = 4.30 · 10−19 J = 2.69 eV (8)

∆VS = Kmax/e = 2.69 V (9)

Problem 28.10. Electrons are ejected from a metallic surface with speeds ranging up to 4.60 · 105 m/s when light with awavelength of 625 nm is used. (a) What is the work function of the surface? (b) What is the cutoff frequency for this surface?

(a) Conserving energy

hf =hc

λ= φ+Kmax = φ+

12mev

2max (10)

φ =hc

λ− 1

2mev

2max =

6.63 · 10−34 J·s · 3.00 · 108 m/s625 · 10−9 m

− 12· 9.11 · 10−31 kg · (4.60 · 105 m/s)2 = 2.21 · 10−19 J = 1.38 eV

(11)

(b) Light at the cutoff frequency only barely supplies enough energy to overcome the work function.

hfcut = φ (12)

fcut =φ

h= 334 THz (13)

(14)

Problem 28.13. An isolated copper sphere of radius 5.00 cm, initially uncharged, is illuminated by ultraviolet light ofwavelength 200 nm. What charge will the photoelectric effect induce on the sphere? The work function for copper is 4.70 eV.

As light lands on the sphere, electrons are blasted off into oblivion. As they leave, the sphere accumulates positive charge, soit takes a bit more energy to pull the next electrons away. Eventually the system reaches equilibrium when the initial kineticenergy of electron blasting off is not quite enough for it to escape the electro-static attraction to the positive sphere. In math

Kmax = hf − φ =hc

λ− φ = 2.40 · 10−19 J = 1.50 eV (15)

= keQq

R(16)

Q =KmaxR

keq=

2.40 · 10−19 J8.99 · 109 N·m2/C2 · 1.60 · 10−19 C

= 8.34 pC = 52.1 · 106 electrons (17)

Problem 28.56. Figure P28.56 shows the stopping potential versus the incident photon frequency for the photoelectric effectfor sodupm. Use the graph to find (a) the work function, (b) the ratio h/e, and (c) the cutoff wavelength. The data are takenfrom R.A. Millikan, Physical Review 7:362 (1916).

0

1

2

3

4

∆V

s(V

)

400 800 1200f (THz)

(a) The fit line passes nearby the points (400 THz, 0 V) and (1.20 PHz, 3.3 V). In point-slope form, the fit line is then

∆Vs − 0 V =(3.3 − 0) V

(1200 − 400) THz(f − 400 THz) (18)

∆Vs = 4.125mV/THz · (f − 400 THz) (19)

The work function is the inverse y-intercept, so

φ = −∆Vs(f = 0) = −4.125mV/THz · (−400 THz) = 1.65 eV (20)

(b) The theoretical form for the fit line is

e∆Vs = hf − φ (21)

∆Vs =h

ef − φ (22)

so he = 4.12 mV/THz = 4.12 pV/s.

(c) The cutoff wavelength is given by the work function and conservation of energy.

hfcut =hc

λcut= φ (23)

λcut =hc

φ= 751 nm (24)

Problem 28.57. The following table shows data obtained in a photoelectric experiment. (a) Using these data, make a graphsimilar to Active Figure 28.9 that plots as a straight line. From the graph, determine (b) an experimental value for Planck’sconstant (in joule-seconds) and (c) the work function (in electron volts) for the surface. (Two significant figures for eachanswer are sufficient.)

Wavelength (nm) Maximum Kinetic Energy of Photoelectrons (eV)588 0.67505 0.98445 1.35399 1.63

(a) We can convert the wavelengths to frequencies and graph them

λ (nm) f = c/λ (THz)588 510505 594445 674399 751

−2

−1

0

1

2

Km

ax

(V)

0 200 400 600 800f (THz)

The least-squares fit yields parameters (c) φ = 1.4 eV and h/e = 4.04 mV/THz. So (b) h = 6.5 · 10−34 J·s.