The Photoelectric Effect and Wave-Particle Duality
- Consider you’re holding a metal plate in a brightly lit room.
- As light shines on the plate, something extraordinary happens: electrons are ejected from its surface.
- But here’s the twist, this only occurs if the light’s frequency is above a certain threshold, no matter how intense the light is.
- Why doesn’t brighter light always eject electrons?
Evidence for Light’s Particle Nature: The Photoelectric Effect
Photoelectric effect
The photoelectric effect refers to the emission of electrons from a metal surface when light or other electromagnetic radiation shines on it.
Scientists study this phenomenon using an experimental setup that includes a photosensitive metal surface, a collecting plate, and a variable voltage supply inside an evacuated tube.
When light strikes the metal, electrons are ejected and travel to the collecting plate, completing an electric circuit and generating a measurable current.
Key Observations of the Photoelectric Effect:
- Threshold Frequency:
- Electrons are emitted only if the light’s frequency exceeds a minimum value, called the threshold frequency ($f_{\text{c}}$).
- Frequency Determines Kinetic Energy:
- The kinetic energy ($E_{\text{max}}$) of the emitted electrons increases with the frequency of the light but is unaffected by its intensity.
- Instantaneous Emission:
- Electrons are emitted immediately after light strikes the surface, with no delay.
- Intensity Affects Quantity, Not Energy:
- Increasing the light’s intensity increases the number of emitted electrons (current) but does not affect their kinetic energy.
Below threshold frequency, no electrons are emitted, regardless of the light’s intensity.
Why Light as a Wave Fails to Explain These Observations
If light were purely a wave:
- Increasing intensity (wave amplitude) would increase the energy of emitted electrons, but this is not observed.
- Electrons would eventually absorb enough energy to escape, even at low frequencies, contradicting the threshold frequency requirement.
- A time delay would occur as electrons accumulate energy from low-intensity light, but emission is instantaneous.
These discrepancies show that the wave model alone cannot explain the photoelectric effect.
Einstein’s Explanation: Light as Photons
- Albert Einstein resolved this dilemma by proposing that light consists of discrete packets of energy called photons.
- The energy ($E$) of a photon is proportional to the frequency ($f$) of the light: $$E = hf$$ where $h$ is Planck’s constant ($6.63 \times 10^{-34} \, \text{J s}$).
- When a photon strikes the metal surface, its energy is transferred to a single electron.
- The electron uses some of this energy to overcome the attractive forces holding it in the metal.
- This minimum energy is called the work function ($\phi$) of the metal.
- Any remaining energy becomes the electron’s kinetic energy ($E_{\text{max}}$): $$E_{\text{max}} = hf - \phi$$
Work function
Work function $\phi$ is the minimum quantity of energy which is required to remove an electron from the surface of a given solid.
Stopping Voltage and Maximum Kinetic Energy
The maximum kinetic energy of the emitted electrons can be measured using the stopping voltage ($V_s$).
Stopping voltage
The stopping voltage is the voltage required to repel all emitted electrons, stopping the current.
The relationship is:
$$eV_s = E_{\text{max}}$$
A metal has a work function of $2.00 \, \text{eV}$. Light with a frequency of $6.00 \times 10^{14} \, \text{Hz}$ shines on it. Calculate:
- he maximum kinetic energy of the emitted electrons.
- The stopping voltage.
Solution
- Photon energy:
$$E = hf = (6.63 \times 10^{-34})(6.00 \times 10^{14}) $$
$$= 3.98 \times 10^{-19} \, \text{J}$$
- Converting to electron volts:
$$E = \frac{3.98 \times 10^{-19} }{ 1.60 \times 10^{-19}} = 2.49 \, \text{eV}$$
- Maximum kinetic energy:
$$E_{\text{max}} = hf - \phi$$
$$ = 2.49 - 2.00 = 0.49 \, \text{eV}$$
- Stopping voltage: $V_s = E_{\text{max}} / e = 0.49 \, \text{V}$.
To convert energy from joules to electron volts, divide by $1.60 \times 10^{-19} \, \text{J eV}^{-1}$.
Threshold Frequency and Work Function
Threshold frequency
The threshold frequency $f_{\text{c}}$ is the minimum frequency of light required to eject electrons.
At this frequency, the photon’s energy equals the work function ($hf_{\text{c}} = \phi$), and the electrons have zero kinetic energy:
$$f_{\text{c}} = \frac{\phi}{h}$$
If the light’s frequency is below $f_{\text{c}}$, no electrons are emitted, regardless of intensity.
What happens to the stopping voltage if the frequency of the light increases? Why?
