Photoelectric effect: If an EM wave with sufficiently high frequency is shone on a piece of metal, electrons will be emitted from the metal surface.
Experiment on photoelectric effect:
Put a UV lamp against the zinc plate on a gold leaf electroscope.
1) When zinc plate is negatively charged, gold leaf fall as UV radiation strikes on it because electrons (negative charge) are emitted away.
2) The gold leaf stopped falling if barrier exist between the lamp and the zinc plate.
3) When zinc plate is positively charged, gold leaf will not fall because electrons can’t escape due to electrostatic attraction between zinc plate (+) and electron (-).
Photocell is composed by a metal plate (cathode) and an electrode (anode). Under exposure of EM wave with sufficiently high frequency, it emits electron. When it’s connected with voltage supply and ammeter, photoelectric current can be measured.
1) Consider anode with a higher potential, the photoelectric effect occurs normally.
2) When anode has a lower potential, the current direction is still the same but the photoelectric current is decreasing.
3) At a certain potential (-Vs) the photoelectric current becomes zero. We say that Vs is the stopping potential of the cell.
4) At stopping potential even electrons with highest K.E. can’t reach the anode. Therefore K.E.max = eVs
5) A smaller unit of energy, electron-volt, 1eV = (e)(V) = 1.6*10-19 J. This is equal to the gain of energy when electron accelerates through a p.d. of 1V.
Properties of photoelectric effect
1) Electrons emitted only when f ≥ f0, the threshold frequency.
2) Number of photoelectrons (per second) is proportional to radiation intensity.
3) K.E.max increases with frequency.
4) Photoelectric effect is immediate, i.e., once radiation with sufficiently high frequency is given to the metal plate, electrons are emitted at once.
Explaining photoelectric effect by classical wave theory:
- Wave energy transmitted in a continuous manner and spreads over the wavefront.
- Energy transfer rate is independent of frequency.
The wave theory cannot explain property 1,3 and 4. Here’s the contradictory result by wave theory on the properties of photoelectric effect:
Property 1: Energy is independent of frequency so it should happen for all frequency.
Property 3: Energy is independent of frequency so as K.E.max.
Property 4: Since energy transfer is continuous, there’s delay before electrons get enough energy to escape.
Quantum theory: quantizing light wave into discrete packets, called light quanta or photons.
The energy of each photon is related to its frequency, E=hf, where h is the Planck constant, which is 6.63 * 10-34 J s.
Note that the behavior of photon is discrete instead of continuous manner.
Since K.E. of photoelectron = energy absorbed – energy used to escape the metal, we have Einstein’s photoelectric equation: K.E.max = hf – Φ, where Φ is the work function, in terms of eV, subjective to the metal used. (usually inversely related to its reactivity.)
The quantum theory can explain most of the photoelectric effect:
Property 1: threshold frequency is given by hf0 = Φ. Therefore we also have K.E.max = h(f – f0).
Property 2: Intensity is proportional to rate of photons transmitted, so it’s also proportional to the photoelectrons emitted.
Property 3: true by K.E.max = hf – Φ.
Property 4: true since electron gain enough energy once it absorb the photon.
Experimental verification of K.E.max = hf – Φ by showing Vs = (h/e) f –Φ/e
Direct a beam of monochromatic light of frequency f, the photoelectrons complete the circuit with voltage supply and galvanometer. Vs is found when the readings of galvanometer drops to zero. A Vs-f graph has x,y-intercept f0 and –Φ/e respectively, and slope h/e. Note that the slope is a constant and applicable to all metal and frequency.
By the above equation we have:
- Vs is independent of intensity, but intensity is proportional to photoelectric current (Ip).
- Under the same intensity, light with higher frequency has larger Vs but lower Ip.
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