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Physics > Quantum Phenomena > Flashcards

Quantum Phenomena Flashcards

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1
Q

What is the photoelectric effect?

A

Electrons are emitted from the surface of a metal when electromagnetic radiation above a certain frequency is directed at the metal.

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2
Q

Describe the layout of the experiment observing the photoelectric effect.

A

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.

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3
Q

When the zinc plate is charged, what happens?

A

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.

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4
Q

What if UV light is directed at the zinc plate?

A

Leaf immediately falls because conduction electrons at zinc surface leave.

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5
Q

What are the emitted electrons referred to as?

A

Photoelectrons.

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6
Q

What if UV light was directed at positively charged plate, and so the electroscope was positive?

A

The leaf would stay risen and would not fall.

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7
Q

What was observed during this experiment? (3)
1 - comment on frequency/λ:

A

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

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8
Q

2- comment on intensity:

A

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.

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9
Q

3- comment on speed of photoelectric emission:

A

Photoelectric emission occurs without delay as soon as em radiation directed at surface, provided f > f0, regardless of intensity.

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10
Q

What can’t the wave theory explain about these observations? What does wave theory say should happen?

A

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.

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11
Q

Einstein explanation of photoelectric effect. (2)

A

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.

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12
Q

What is the work function?
What happens to excess energy from photon?

A

The minimum energy needed by an e- to escape from the metal surface at zero potential.
Excess energy gained by the photoelectron becomes Ek.

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13
Q

What’s the maximum kinetic energy that the photoelectron can have?
Hence, what is f0?

A

Ekmax = hf - Φ ∴ hf = Ekmax + Φ
Emission can take place provided Ek max > 0 or hf > Φ.
Hence, f0 = Φ/h

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14
Q

(Subtopic: stopping voltage)

A

Electrons that escape from the metal plate can be attracted back to it by giving the plate a sufficiently +ve charge.

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15
Q

What is stopping voltage, Vs?

A

The minimum potential needed to stop photoelectric emission.

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16
Q

Comment on Ekmax of emitted e-s at this voltage.

Relate Ekmax to Vs.

A

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)

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17
Q

Energy of each vibrating atom is quantised. What does this mean?

A

Only certain levels of energy are allowed.

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18
Q

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)

A

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.

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19
Q

What about if the e- doesn’t leave the metal?

A

If the e- doesn’t leave, it collides repeatedly with other e-s and positive ions, and it quickly loses its extra Ek.

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20
Q

Describe vacuum photocell setup.

A

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.

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21
Q

What is the photoelectric current?

A

The number of photoelectrons per second that transfer from cathode to anode.

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22
Q

Photoelectric current is proportional to intensity of radiation incident on surface of cathode. Why?

A

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.

23
Q

Intensity of light doesn’t affect Ekmax of e-. Why?

A

Energy gained is due to absorption of one photon only.

24
Q

Ekmax of e-s emitted for a given frequency can be measured using a photocell.

A

Ekmax = eVs
Here, current = 0

25
Q

Graph of Ekmax plotted against diff values of f, a straight line forms. What’s the equation of that line?

A

Ekmax = hf - Φ
y = mx + c

26
Q

How do we measure ionisation energy?

A

Make electron collide at increasing speed with gas atoms in a sealed tube.

27
Q

Describe layout of experiment.

A

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.

28
Q

Why do we use gas at low pressure?

A

Otherwise there’ll be too many atoms in the tube and so the electrons can’t reach the anode.

29
Q

How do we increase the speed of the electrons?

A

Increase the p.d. across the filament and anode.

30
Q

Why are we increasing speed?

A

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.

31
Q

What do we observe when ionisation has occurred?

A

Ionisation near anode causes much greater current to pass through ammeter.

32
Q

How do we work out the ionisation energy of each gas atom from this?

A

IE = work done on each e- from filament (converted to Ek) = eV

33
Q

Using gas-filled tubes with metal grid between filament and anode, what can we show about the gas atoms?

A

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.

34
Q

What happens to current during excitation?

A

If the e- loses all its Ek when it causes excitation , current is reduced.

35
Q

What happens if colliding e- doesn’t have enough Ek to cause excitation?

A

It’s deflected by the atom, with no overall loss in Ek.

36
Q

What does excitation energies mean?
How do we measure these?

A

Energy values at which an atom absorbs energy.
Can measure by increasing p.d. and measuring p.d. when current falls.

37
Q

Why is energy needed for excitation to occur?

A

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.

38
Q

What’s the ground state?

A

The lowest energy state of an atom.

39
Q

Excitation using photons: what’s the condition?

A

Photons must be of energy exactly equal to the difference in energy level.

40
Q

Certain substances fluoresce or glow with visible light when they absorb UV radiation. How?

A

Atoms absorb UV photons, are excited, de-excite, emit visible photons (of longer wavelength).
When remove UV source, they stop glowing.

41
Q

Explain how a fluorescent tube works? (inc description of what it is)

A

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.

42
Q

Is this more or less efficient than a filament lamp?

A

This is more efficient because less energy is wasted as thermal energy.

43
Q

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?

A

By measuring the wavelengths of a line spectrum, we can ∴ identify the element that produced the light.

44
Q

No other element produced the same pattern of light wavelengths. How do we know this?

A

This is bc the energy levels of each type of atom are unique to that atom.

45
Q

Wave-particle duality of light: what experiment confirms each?

A

Wave-like nature is observed when diffraction of light takes place.
Particle-like nature observed in photoelectric effect.

46
Q

What is evidence for particle-like nature of electrons?

A

Electrons in beam can be deflected by a magnetic filed.

47
Q

Matter waves - de Broglie hypothesis?
(talk abt λ)

A

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.

48
Q

Evidence of hypothesis explained fully (e-s can be diffracted)?

A

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.)

49
Q

How do we increase the speed of the e-s, and what affect does this have?

A

Increase speed by increasing p.d. between filament and metal plate. This makes diffraction ring smaller bc increasing v decreases λ which decreases diffraction.

50
Q

Why does an e- in an atom have a fixed amount of energy that depends on the shell it occupies?

A

Its de Broglie λ has to fit the shape and size of the shell.

51
Q

TEM - relate how detailed image is produced to this topic.

Clue - about electrons as waves.

A

e-s accelerated ∴ v increases ∴ λ decreases ∴ very detailed image produced.

52
Q

MRI - how it works

Clue - excited/de-excited H

A

detect radio waves emitted when H atoms flip between energy levels in strong magnetic field.

53
Q

SQIDs - what does it do?

A

detect very very weak magnetic fields eg those produced by electrical activity in brain.

Physics (29 decks)

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