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  4. Photons and matter waves
  5. Photoelectric effect

6.2 Photoelectric effect  (Page 5/17)

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Work function and cut-off frequency

When a 180-nm light is used in an experiment with an unknown metal, the measured photocurrent drops to zero at potential – 0.80 V. Determine the work function of the metal and its cut-off frequency for the photoelectric effect.

Strategy

To find the cut-off frequency f c , we use [link] , but first we must find the work function ϕ . To find ϕ , we use [link] and [link] . Photocurrent drops to zero at the stopping value of potential, so we identify Δ V s = 0.8 V .

Solution

We use [link] to find the kinetic energy of the photoelectrons:

K max = e Δ V s = e ( 0.80 V ) = 0.80 eV .

Now we solve [link] for ϕ :

ϕ = h f − K max = h c λ − K max = 1240 eV · nm 180 nm − 0.80 eV = 6.09 eV .

Finally, we use [link] to find the cut-off frequency:

f c = ϕ h = 6.09 eV 4.136 × 10 −15 eV · s = 1.47 × 10 −15 Hz .

Significance

In calculations like the one shown in this example, it is convenient to use Planck’s constant in the units of eV · s and express all energies in eV instead of joules.

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The photon energy and kinetic energy of photoelectrons

A 430-nm violet light is incident on a calcium photoelectrode with a work function of 2.71 eV.

Find the energy of the incident photons and the maximum kinetic energy of ejected electrons.

Strategy

The energy of the incident photon is E f = h f = h c / λ , where we use f λ = c . To obtain the maximum energy of the ejected electrons, we use [link] .

Solution

E f = h c λ = 1240 eV · nm 430 nm = 2.88 eV , K max = E f − ϕ = 2.88 eV − 2.71 eV = 0.17 eV

Significance

In this experimental setup, photoelectrons stop flowing at the stopping potential of 0.17 V.

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Check Your Understanding A yellow 589-nm light is incident on a surface whose work function is 1.20 eV. What is the stopping potential? What is the cut-off wavelength?

−0.91 V; 1040 nm

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Check Your Understanding Cut-off frequency for the photoelectric effect in some materials is 8.0 × 10 13 Hz . When the incident light has a frequency of 1.2 × 10 14 Hz , the stopping potential is measured as – 0.16 V. Estimate a value of Planck’s constant from these data (in units J · s and eV · s ) and determine the percentage error of your estimation.

h = 6.40 × 10 −34 J · s = 4.0 × 10 −15 eV · s; − 3.5 %

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Summary

  • The photoelectric effect occurs when photoelectrons are ejected from a metal surface in response to monochromatic radiation incident on the surface. It has three characteristics: (1) it is instantaneous, (2) it occurs only when the radiation is above a cut-off frequency, and (3) kinetic energies of photoelectrons at the surface do not depend of the intensity of radiation. The photoelectric effect cannot be explained by classical theory.
  • We can explain the photoelectric effect by assuming that radiation consists of photons (particles of light). Each photon carries a quantum of energy. The energy of a photon depends only on its frequency, which is the frequency of the radiation. At the surface, the entire energy of a photon is transferred to one photoelectron.
  • The maximum kinetic energy of a photoelectron at the metal surface is the difference between the energy of the incident photon and the work function of the metal. The work function is the binding energy of electrons to the metal surface. Each metal has its own characteristic work function.
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Source:  OpenStax, University physics volume 3. OpenStax CNX. Nov 04, 2016 Download for free at http://cnx.org/content/col12067/1.4
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