bei48482_FM

E = hv 0

Metal

E = hv

KE = 0

KEmax = hv – hv 0

Figure 2.13If the energy h 0 (the work function of the surface) is needed to remove an electron from

a metal surface, the maximum electron kinetic energy will be hh 0 when light of frequency is

directed at the surface.

electron to leave the metal. Because h 0 , Eq. (2.8) can be rewritten (Fig. 2.13)

hKEmaxh 0

KEmaxhh 0 h( 0 ) (2.9)

This formula accounts for the relationships between KEmaxand plotted in Fig. 2.12

from experimental data. If Einstein was right, the slopes of the lines should all be equal

to Planck’s constant h, and this is indeed the case.

In terms of electronvolts, the formula E hfor photon energy becomes

E(4.136 10 ^15 )eVs (2.10)

If we are given instead the wavelength of the light, then since cwe have

E

(2.11)

Example 2.2

Ultraviolet light of wavelength 350 nm and intensity 1.00 W/m^2 is directed at a potassium sur-

face. (a) Find the maximum KE of the photoelectrons. (b) If 0.50 percent of the incident pho-

tons produce photoelectrons, how many are emitted per second if the potassium surface has an

area of 1.00 cm^2?

Solution

(a) From Eq. (2.11) the energy of the photons is, since 1 nm 1 nanometer  10 ^9 m,

Ep3.5 eV

1.24  10 ^6 eVm



(350 nm)(10^9 m/nm)

1.240 10 ^6 eVm





(4.136 10 ^15 eVs)(2.998 108 m/s)





Photon

energy

6.626 10 ^34 Js



1.602 10 ^19 J/eV

Photon

energy

66 Chapter Two

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