The photoelectric effect
| English | Chinese | Pinyin |
|---|---|---|
| photoelectric effect | 光电效应 | guāng diàn xiào yìng |
| photoelectrons | 光电子 | guāng diàn zi |
| threshold frequency | 极限频率 | jí xiàn pín lǜ |
| work function | 逸出功 | yì chū gōng |
| threshold wavelength | 极限波长 | jí xiàn bō cháng |
| stopping potential | 遏止电势 | è zhǐ diàn shì |
The brightest red lamp will not do it
- Shine a blazing red lamp on a clean zinc plate and nothing comes off, however long you wait.
- Swap it for a feeble ultraviolet lamp and electrons are emitted at once, from the very first instant.
- A wave carries more energy when it is brighter, so a wave model says the red lamp should eventually win. It never does.
- That single stubborn fact is why photons exist. This lesson is the photoelectric effect 光电效应 and the equation that explains it.
What the effect is
- The photoelectric effect is the emission of electrons from the surface of a metal when electromagnetic radiation of high enough frequency is incident on it.
- Two marks: emission of electrons, and from a metal surface illuminated by electromagnetic radiation. The emitted electrons are photoelectrons 光电子.
- The classic demonstration is a negatively charged zinc plate on a gold-leaf electroscope: shine ultraviolet on it and the leaf falls as the charge leaks away.
The leaf drops only under ultraviolet
Threshold frequency and work function
- Every metal has a threshold frequency 极限频率 $f_0$: the minimum frequency of radiation for which photoelectrons are emitted. Below it, nothing, however bright.
- The work function 逸出功 $\Phi$ is the minimum energy needed to remove an electron from the surface of the metal, and
- Both "minimum" and "from the surface" carry marks. An electron deeper inside needs more, which is exactly why the equation gives a maximum kinetic energy.
- The threshold wavelength 极限波长 $\lambda_0 = hc/\Phi$ is a maximum: longer wavelengths do nothing at all. Typical work functions are $2$ to $5\ \text{eV}$.
Below the threshold frequency, shining a brighter light on the metal:
Each photon is below $\Phi$, so no single photon can free an electron — brightness cannot help.
The minimum energy needed to free an electron from the surface is the ____ function.
The work function $\Phi = hf_0$ — the least energy to release a surface electron.
Match each term to the definition the examiner marks.
Frequency has a minimum, wavelength a maximum: they run opposite ways. "Minimum" and "from the surface" both carry marks in the work function.
Worked example: which metals emit
- Light of $400\ \text{nm}$ falls on caesium ($2.1\ \text{eV}$), sodium ($2.3\ \text{eV}$), zinc ($4.3\ \text{eV}$) and platinum ($5.6\ \text{eV}$). Which emit, and with what maximum kinetic energy?
- Photon energy: $hc/\lambda = 1240/400 = 3.1\ \text{eV}$.
- Emission needs $hf \geq \Phi$. So caesium emits, with $3.1 - 2.1 = 1.0\ \text{eV}$, and sodium with $0.8\ \text{eV}$.
- Zinc and platinum emit nothing, however intense the light. To make zinc emit you need $\lambda < 1240/4.3 = 290\ \text{nm}$, in the ultraviolet, which is why the electroscope demonstration uses a UV lamp on zinc.
Light of 400 nm (photon energy 3.1 eV) falls on these metals. Which emit photoelectrons? Select all that apply.
Emission needs hf >= work function. Zinc and platinum emit nothing at this wavelength however intense the light; zinc needs below 290 nm, in the ultraviolet.
Einstein's photoelectric equation
- One photon gives all its energy to one electron. There is no sharing and no saving up.
- If $hf$ exceeds the work function, the electron escapes with kinetic energy up to a maximum:
- So the maximum kinetic energy depends linearly on frequency, and not at all on brightness.
The work function first, the rest as speed
A $5.0\ \text{eV}$ photon hits a metal with work function $2.0\ \text{eV}$. What is the maximum KE of the photoelectron?
Max KE $= hf - \Phi = 5.0 - 2.0 = 3.0\ \text{eV}$.
The maximum KE of photoelectrons depends on the frequency, not the brightness.
Max KE $= h(f - f_0)$ — set by frequency alone. Brightness changes the number of electrons.
Worked example: work function and maximum speed
- Magnesium emits only above $8.8\times10^{14}\ \text{Hz}$. It is illuminated at $1.2\times10^{15}\ \text{Hz}$. Find the work function and the maximum speed of the photoelectrons.
- $\Phi = hf_0 = (6.63\times10^{-34})(8.8\times10^{14}) = 5.8\times10^{-19}\ \text{J} = 3.6\ \text{eV}$.
- $\tfrac{1}{2}mv_{\text{max}}^2 = h(f - f_0) = (6.63\times10^{-34})(1.2\times10^{15} - 8.8\times10^{14}) = 2.1\times10^{-19}\ \text{J}$.
- $v_{\text{max}} = \sqrt{\dfrac{2\times 2.1\times10^{-19}}{9.11\times10^{-31}}} = 6.8\times10^{5}\ \text{m/s}$.
- Subtract the frequencies before multiplying by $h$. Rounding $hf$ and $\Phi$ separately and then subtracting loses a significant figure in the small difference.
A metal has threshold frequency 8.8e14 Hz and is lit at 1.2e15 Hz. What is the maximum kinetic energy of the photoelectrons, in units of 1e-19 J?
h(f - f0) = 6.63e-34 x 3.2e14 = 2.1e-19 J. Subtract the frequencies FIRST; rounding hf and the work function separately loses a figure in the small difference.
The straight-line graph
- Write the equation as $E_{\text{K,max}} = hf - \Phi$. That is a straight line with gradient $h$, intercept $-\Phi$ on the energy axis, and $f_0 = \Phi/h$ on the frequency axis.
- The gradient is the same for every metal, since $h$ is a constant. So two metals give two parallel lines, the larger work function cutting the frequency axis further to the right.
- The intensity of the light moves neither line. "Sketch the line for metal Y" is marked on exactly those two features: parallel, and shifted.
- This is how the Planck constant is measured.
Same slope, different intercepts
The photoelectric effect
KEmax = h·f − φ
Max KE is a straight line in frequency, with intercept −φ (the work function).
On a graph of maximum kinetic energy against frequency, the gradient equals the ____ constant, so the lines for two different metals are parallel.
E = hf - Phi, so gradient h and intercept -Phi. The metal changes only the intercepts, never the slope, and the intensity changes neither.
Measuring it: the photocell
- Photoelectrons cross a vacuum to a collector, and the current counts electrons per second.
- Make the collector negative so the electrons must climb a potential hill. Raise the reverse p.d. until the current just reaches zero: that is the stopping potential 遏止电势 $V_{\text{s}}$, and
Turn the voltage up until the count reaches zero
Put the stopping-potential measurement in order.
The current counts electrons per second; the stopping potential measures the energy of the fastest one. Intensity moves the first, frequency the second.
Intensity versus frequency
- At fixed frequency, doubling the intensity doubles the current but leaves the stopping potential unchanged.
- Raising the frequency raises the stopping potential but, at fixed intensity, does not raise the current.
- The reason is the one-photon-one-electron rule: intensity sets how many photons arrive per second, so it sets how many electrons leave. Frequency sets how much energy each one carries, so it sets how fast they leave.
- Keep those two sentences apart in an answer. Most lost marks here come from writing one when the question asked for the other.
Doubling the brightness at the same frequency:
More photons per second → more electrons (more current), but each photon still carries the same $hf$.
Why a wave model cannot do it
- A wave model predicts brightness should set the electrons' energy, and that any frequency should work given enough time to accumulate energy. Three observations kill it.
- No emission below $f_0$, however bright and however long you wait.
- Immediate emission at or above $f_0$, even from a very dim source.
- Maximum kinetic energy depends on frequency, not on brightness.
- All three follow at once from one photon delivering $hf$ to one electron, whole.
The wave model wrongly predicts that any frequency would emit electrons if you wait long enough.
Waves would let energy build up gradually — but experiment shows a sharp threshold frequency, which the photon model explains.
You've got it
- the photoelectric effect is the emission of electrons from a metal surface illuminated by radiation of high enough frequency
- the work function is the minimum energy to remove an electron from the surface, and $\Phi = hf_0$
- $hf = \Phi + \tfrac{1}{2}mv_{\text{max}}^2$, so $E_{\text{K,max}}$ against $f$ is a straight line of gradient $h$ and intercept $-\Phi$, parallel for every metal
- intensity sets the current, frequency sets the maximum kinetic energy, and the stopping potential gives $eV_{\text{s}} = E_{\text{K,max}}$