Quantum Phenomena Flashcards
What is the photoelectric effect?
Electrons are emitted from the surface of a metal when electromagnetic radiation above a certain frequency is directed at the metal.
Describe the layout of the experiment observing the photoelectric effect.
UV radiation is directed at a negatively charged zinc plate attached to a gold leaf electroscope. This device is a very sensitive detector of charge.
When the zinc plate is charged, what happens?
Thin gold leaf of electroscope rises and stays in that position - it’s repelled from the metal stem bc they have the same type of charge.
What if UV light is directed at the zinc plate?
Leaf immediately falls because conduction electrons at zinc surface leave.
What are the emitted electrons referred to as?
Photoelectrons.
What if UV light was directed at positively charged plate, and so the electroscope was positive?
The leaf would stay risen and would not fall.
What was observed during this experiment? (3)
1 - comment on frequency/λ:
Photelectric emission of e-s won’t take place if f of incident em radiation is below threshold frequency. f0 depends on type of metal.
∴ λ must be < λmax = c/f0
2- comment on intensity:
Number of e-s emitted per second is proportional to intensity, provided f > f0. If f< f0 no photoelectric effect takes place no matter how high the intensity of the incident radiation.
3- comment on speed of photoelectric emission:
Photoelectric emission occurs without delay as soon as em radiation directed at surface, provided f > f0, regardless of intensity.
What can’t the wave theory explain about these observations? What does wave theory say should happen?
Can’t explain existence of f0 or why photoelectric effect occurs without delay.
Wave theory says each conduction e- at the surface of the metal should gain some energy from the incoming waves, regardless of how many waves arrive each second.
Einstein explanation of photoelectric effect. (2)
When light is incident on a metal surface, an e- at the surface absorbs a single photon ∴ gains E = hf. One-to-one interaction.
An e- can leave the metal surface if the energy gained from a single photon > work function of the metal.
What is the work function?
What happens to excess energy from photon?
The minimum energy needed by an e- to escape from the metal surface at zero potential.
Excess energy gained by the photoelectron becomes Ek.
What’s the maximum kinetic energy that the photoelectron can have?
Hence, what is f0?
Ekmax = hf - Φ ∴ hf = Ekmax + Φ
Emission can take place provided Ek max > 0 or hf > Φ.
Hence, f0 = Φ/h
(Subtopic: stopping voltage)
Electrons that escape from the metal plate can be attracted back to it by giving the plate a sufficiently +ve charge.
What is stopping voltage, Vs?
The minimum potential needed to stop photoelectric emission.
Comment on Ekmax of emitted e-s at this voltage.
Relate Ekmax to Vs.
At this voltage Ekmax is reduced to zero bc each emitted e- must do extra work equal to e x Vs to leave the metal surface.
Ekmax = eVs
(can use photocell to measure Ekmax by measuring Vs, this is when current = 0)
Energy of each vibrating atom is quantised. What does this mean?
Only certain levels of energy are allowed.
Conduction e-s in a metal move about at random. The average Ek of a conduction e- depends on the temp of the metal.
When a conduction e- absorbs a photon…. (think energy transfers)
its Ek increases by an amount equal to the energy of the photon. if Ek > Φ of metal, the conduction e- can leave the metal.
What about if the e- doesn’t leave the metal?
If the e- doesn’t leave, it collides repeatedly with other e-s and positive ions, and it quickly loses its extra Ek.
Describe vacuum photocell setup.
A glass tube that contains a metal plate (photocathode) and a smaller metal electrode (anode). When light of f> f0 directed at photocathode, e-s emitted from cathode and attracted to anode. Micrometer in circuit used to measure photoelectric current.
What is the photoelectric current?
The number of photoelectrons per second that transfer from cathode to anode.
Photoelectric current is proportional to intensity of radiation incident on surface of cathode. Why?
Photoelectric current ∝
Light intensity is a measure of energy/sec carried by light ∝ number of photons per second incident on cathode ∝ number of electrons emitted per second, because 1-to-1 interaction.
Intensity of light doesn’t affect Ekmax of e-. Why?
Energy gained is due to absorption of one photon only.
Ekmax of e-s emitted for a given frequency can be measured using a photocell.
Ekmax = eVs
Here, current = 0
Graph of Ekmax plotted against diff values of f, a straight line forms. What’s the equation of that line?
Ekmax = hf - Φ
y = mx + c
How do we measure ionisation energy?
Make electron collide at increasing speed with gas atoms in a sealed tube.
Describe layout of experiment.
Electrons are emitted from a heated filament (thermionic emission) in the tube and are attracted to the +ve metal plate (anode) at the other end. Gas at low pressure in tube. Ammeter records very small current due to e-s from filament.
Why do we use gas at low pressure?
Otherwise there’ll be too many atoms in the tube and so the electrons can’t reach the anode.
How do we increase the speed of the electrons?
Increase the p.d. across the filament and anode.
Why are we increasing speed?
No ionisation until electrons from filament reach certain speed. At this speed each e- arrives at anode with just enough Ek to ionise a gas atom by knocking an e- out of the atom.
What do we observe when ionisation has occurred?
Ionisation near anode causes much greater current to pass through ammeter.
How do we work out the ionisation energy of each gas atom from this?
IE = work done on each e- from filament (converted to Ek) = eV
Using gas-filled tubes with metal grid between filament and anode, what can we show about the gas atoms?
We can show gas atoms can absorb energy from colliding e-s without being ionised. This is excitation, and happens at certain energies which are characteristic of the atoms of the gas.
What happens to current during excitation?
If the e- loses all its Ek when it causes excitation , current is reduced.
What happens if colliding e- doesn’t have enough Ek to cause excitation?
It’s deflected by the atom, with no overall loss in Ek.
What does excitation energies mean?
How do we measure these?
Energy values at which an atom absorbs energy.
Can measure by increasing p.d. and measuring p.d. when current falls.
Why is energy needed for excitation to occur?
When excitation occurs, colliding e- makes atomic e- move to a higher shell/energy level. Energy is needed for this process bc atomic e- moves away from the nucleus.
What’s the ground state?
The lowest energy state of an atom.
Excitation using photons: what’s the condition?
Photons must be of energy exactly equal to the difference in energy level.
Certain substances fluoresce or glow with visible light when they absorb UV radiation. How?
Atoms absorb UV photons, are excited, de-excite, emit visible photons (of longer wavelength).
When remove UV source, they stop glowing.
Explain how a fluorescent tube works? (inc description of what it is)
Fluorescent tube is a glass tube with fluorescent coating on inner surface. Electrons excite mercury atoms. De-excite and emit UV photons, (visible photons and other photons). UV photons excite coating atoms. De-excite and emit visible photons.
Is this more or less efficient than a filament lamp?
This is more efficient because less energy is wasted as thermal energy.
The wavelengths of the lines of a line spectrum of an element are characteristic of the atoms of that element. What does this mean practically?
By measuring the wavelengths of a line spectrum, we can ∴ identify the element that produced the light.
No other element produced the same pattern of light wavelengths. How do we know this?
This is bc the energy levels of each type of atom are unique to that atom.
Wave-particle duality of light: what experiment confirms each?
Wave-like nature is observed when diffraction of light takes place.
Particle-like nature observed in photoelectric effect.
What is evidence for particle-like nature of electrons?
Electrons in beam can be deflected by a magnetic filed.
Matter waves - de Broglie hypothesis?
(talk abt λ)
Matter particles have a dual wave-particle nature too.
Wave-like behaviour of matter particles characterised by de Broglie’s wavelength, λb which is related to the momentum of the particle by λb = h/p = h/mv.
λb can be altered by changing v.
Evidence of hypothesis explained fully (e-s can be diffracted)?
Electrons from heated filament wire (thermionic emission) attracted to a positive metal plate, which has a small hole at its centre. Electrons that pass through the hole form a narrow beam.
Beam directed at thin metal foil, which causes the beam to be diffracted.
Electrons are diffracted in certain directions only. Form a pattern of rings on a fluorescent screen at the end of the tube. Each ring is due to e-s diffracted by the same angle.
(Each ring is the bright part and the gaps between them is the dark fringes.)
How do we increase the speed of the e-s, and what affect does this have?
Increase speed by increasing p.d. between filament and metal plate. This makes diffraction ring smaller bc increasing v decreases λ which decreases diffraction.
Why does an e- in an atom have a fixed amount of energy that depends on the shell it occupies?
Its de Broglie λ has to fit the shape and size of the shell.
TEM - relate how detailed image is produced to this topic.
Clue - about electrons as waves.
e-s accelerated ∴ v increases ∴ λ decreases ∴ very detailed image produced.
MRI - how it works
Clue - excited/de-excited H
detect radio waves emitted when H atoms flip between energy levels in strong magnetic field.
SQIDs - what does it do?
detect very very weak magnetic fields eg those produced by electrical activity in brain.
