06.07 The photoelectric effect

Module 06 โ€” Heat and Particles ยท quiz worth 15 pts

The big idea One photon transfers energy hf to one electron. The material's work function sets a threshold frequency; above threshold, frequency controls maximum electron energy while intensity mainly controls how many electrons are emitted.

Threshold first

Every photoelectric question starts with whether one photon has enough energy to free an electron.

Not sure about the basics underneath this lesson?
Before you start Review photon energy E = hf, frequency, and electronvolts from 06.06. Be ready to read slope and intercept on a straight-line graph.

What the experiment measures

Light striking a suitable material can eject electrons. The key result is a threshold frequency: below it, no electrons are emitted even if the light is intense. Above it, emission can begin without the delay predicted by a purely classical energy-spreading model. Einstein modeled the exchange as one photon with energy hf interacting with one electron.

The energy budget

The work function ฯ† is the minimum energy required to free an electron from the surface. A photon supplies hf. Any remaining energy can become the emitted electron's maximum kinetic energy. Real emitted electrons may have less than the maximum because they begin in different states or lose energy before escaping.

Photoelectric relations

Maximum kinetic energy
Kmaxโก=hfโˆ’ฯ•K_{\max}=hf-\phi

Emission requires hf โ‰ฅ ฯ†.

Threshold
ฯ•=hf0\phi=hf_0

At threshold, Kmax = 0.

Stopping potential
Kmaxโก=eVsK_{\max}=eV_s

e is the elementary charge magnitude, 1.602 ร— 10โปยนโน C.

Electronvolt
1โ€‰eV=1.602ร—10โˆ’19โ€‰J1\,\text{eV}=1.602\times10^{-19}\,\text{J}

Use eV consistently with h = 4.136 ร— 10โปยนโต eVยทs when convenient.

Frequency versus intensity

Check the threshold first

If f < fโ‚€, no photoelectrons are emitted, regardless of intensity in the ideal model.

Change frequency above threshold

Higher f gives each photon more energy, so Kmax and stopping potential increase.

Change intensity above threshold

At fixed frequency, greater intensity means a larger photon rate. It can increase emitted-electron count and current, but not Kmax.

Keep material fixed

The work function and threshold belong to the material and surface condition.

Kmax versus frequency graph: no emission below f0, a physical line rising with slope h from the course graph's approximate f0 of 3.8 ร— 10ยนโด Hz, a dashed extrapolation to an approximate โˆ’ฯ† of โˆ’1.66 eV, and an interpolated Kmax of about 2.57 eV at 10 ร— 10ยนโด Hz.
The measured branch begins at (fโ‚€, 0) and rises with slope h. Its dashed mathematical extension reaches โˆ’ฯ†; below fโ‚€ there is no emission, not a zero-KE plateau. The labeled numbers are approximate readings from the supplied course graph.
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Read the photoelectric graph

On a graph of maximum kinetic energy versus light frequency, the physical data begin at the threshold point (fโ‚€, 0). Above threshold, the points follow a straight line with slope h. Extending that mathematical line backward to frequency zero gives a vertical intercept of โˆ’ฯ†, even though the below-threshold region does not represent emitted electrons with negative kinetic energy.

Worked photoelectric example

Light on a metal surface

  • Frequency. f=5.00ร—1014โ€‰Hzf=5.00\times10^{14}\,\text{Hz}
  • Work function. ฯ•=1.62โ€‰eV\phi=1.62\,\text{eV}
  • Planck constant. h=4.136ร—10โˆ’15โ€‰eVโ‹…sh=4.136\times10^{-15}\,\text{eVยทs}

Find photon energy, maximum kinetic energy, and threshold frequency.

  1. Find photon energy.
    E=hf=(4.136ร—10โˆ’15)(5.00ร—1014)=2.07โ€‰eVE=hf=(4.136\times10^{-15})(5.00\times10^{14})=2.07\,\text{eV}
  2. Subtract the work function.
    Kmaxโก=2.07โˆ’1.62=0.45โ€‰eVK_{\max}=2.07-1.62=0.45\,\text{eV}
  3. Find threshold frequency.
    f0=ฯ•h=1.624.136ร—10โˆ’15=3.92ร—1014โ€‰Hzf_0=\frac{\phi}{h}=\frac{1.62}{4.136\times10^{-15}}=3.92\times10^{14}\,\text{Hz}

E = 2.07 eV, Kmax = 0.45 eV, and fโ‚€ โ‰ˆ 3.92 ร— 10ยนโด Hz.

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Photoelectric traps

Intensity cannot rescue sub-threshold light

More photons of insufficient energy still do not individually overcome the work function in the standard one-photon effect.

Current and maximum energy are different

Intensity can raise current above threshold; frequency raises Kmax.

The y-intercept is negative

For Kmax = hf โˆ’ ฯ†, the extended line crosses the vertical axis at โˆ’ฯ†, not +ฯ†.

Use one energy unit system

Do not mix h in Jยทs with ฯ† in eV without conversion.

Photoelectric practice

First compare f with fโ‚€. Only then calculate an emitted electron's energy.