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A-Level Physics > Quantum Physics > Flashcards

Quantum Physics Flashcards

1
Q

How does electromagnetic energy exsist?

A

Only in certain values - it appeared to come in little packets/ quantas

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2
Q

What is the photon model?

A

EM radiation comes in tiny packets of energy, rather than one continuous wave

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3
Q

What are the packets that EM radiation come in called?

A

Photons

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4
Q

What are the equations for the energy of a photon?

A

E = hf
E = hc /λ

E = energy of photon J
h = Plancks constant
f = frequency Hz
c = speed of light 3 x 10^8 ms^-1

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5
Q

What is the value of Plancks constant?

A

6.63 x 10^-34 Js

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6
Q

At subatomic level, what unit for energy is used?

A

The electronvolt

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7
Q

What is 1 electron volt equal to?

A

1.6 x 10^-19 J

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8
Q

How do you convert from electron volts to joules?

A

multiply by 1.6 x 10^19

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9
Q

What happens when a large enough potential difference is applied across an LED?

A

It emits photons that all the have the same wavelength and frequency.

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10
Q

When an LED just begins to glow, what is the energy lost by each electron converted to?

A

The energy of a single photon

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11
Q

What is the equation for energy lost by each electron?

A

E = eV

E = energy lost by electron
e = elementary charge
V = potential difference against the LED

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12
Q

What is the equation for energy lost by electron = energy of photon?

A

eV = hc/λ

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13
Q

What equation can be used to estimate Plancks constant?

A

eV = hc/λ

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14
Q

How do you determine Plancks constant using LEDs?

A
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15
Q

What is the photoelectric effect?

A

The phenomena in which electrons are emitted from the surface of a metal upon the absorption of electromagnetic radiation

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16
Q

What are the electrons emitted during the photoelectric effect known as?

A

Photoelectrons

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17
Q

What does the photoelectric effect provide evidence for?

A

The particle nature of light

18
Q

How is the photoelectric effect shown with the gold leaf electroscope?

A
  • Briefly touching the top plate with the negative electrode from a high-voltage power supply will charge the electroscope
  • Excess electrons are deposited onto the plate and stem of the electroscope
  • Any charge developed on the plate at the top of the electroscope spreads to the stem and the gold leaf
  • As both the stem and gold leaf have the same charge, they repel each other, causing the leaf to lift away from the stem
  • If UV radiation is then shone onto the zinc surface, the the leaf will slowly fall back towards the stem
  • This shows that the electroscope has gradually lost its negative charge because the incident radiation has caused the free electrons to be emitted from the zinc plate
19
Q

What are the 5 main observations of the photoelectric effect?

A
  • Placing the UV radiation source closer to the metal plate causes the gold leaf to fall more quickly
  • Using a higher frequency light source does not change how quickly the gold leaf falls
  • Using a filament light source causes no charge in the gold leafs position
  • Using a positively charged plate causes no change in the gold leafs position
  • Emission of photoelectrons happens as soon as the radiation is incident on the surface of the metal
20
Q

What is the explanation for the observation ‘Placing the UV radiation source closer to the metal plate causes the gold leaf to fall more quickly’?

A
  • Placing the UV source closer to the plate increases the intensity of incident radiation on the surface of the metal
  • Increasing the intensity, or brightness, of the incident radiation increases the number of photoelectrons emitted per second
  • Therefore, the gold leaf loses negative charge quicker
21
Q

What is the explanation of the observation ‘Using a higher frequency light source does not change how quickly the gold leaf falls ‘?

A
  • The maximum kinetic energy of the emitted electrons increases with the frequency of the incident radiation
  • In the case of the photoelectric effect, energy and frequency are independent of the intensity of the radiation
  • So, the intensity of the incident radiation effects how quickly the gold leaf falls, not the frequency
22
Q

What is the explanation of the observation ‘Using a filament light source causes no charge in the gold leafs position’ ?

A
  • If the incident frequency is below a certain threshold no electrons are emitted, no matter the intensity of the radiation
  • A filament lamp source has a frequency below the threshold frequency of the metal, so no photoelectrons are released
23
Q

What is the explanation of the observation ‘Using a positively charged plate causes no change in the gold leafs position’ ?

A
  • If the plate is positively charged that means there is an excess of positive charge on the plate
  • Electrons are negatively charged so they will not be emitted unless they are on the surface of the metal
  • Any electrons emitted will be attracted back by positive charges on the surface of the metal
24
Q

What is the explanation of the observation ‘Emission of photoelectrons happens as soon as the radiation is incident on the surface of the metal’?

A
  • A single photon interacts with a single electron
  • If the energy of the photon is equal to the work function of the metal, photoelectrons will be released instantaneously
25
Q

Explain the one to one interaction of photons and surface electrons

A

A single photon delivers its energy to a single surface electron. Each surface electron can ONLY interact with a single photon. This means the number of photoelectrons emitted is exactly equal to the number of photons incident on the surface in which the photoelectric effect is taking place

26
Q

What is the energy of a single photon equal to?

A

Minimum energy required to free a single electron from the metal surface + maximum kinetic energy of the emitted electron

27
Q

What is the work function and kinetic energy equation?

A

hf = Φ + KEmax

h = Plancks constant
f = frequency Hz
Φ = work function J
KEmax = maximum kinetic energy

28
Q

What does the work function and KEmax equation show?

A
  • If the incident photons do not have high enough frequency and energy to overcome the work function, then no electrons will be emitted
  • hf0 = Φ where f0 = threshold frequency; here photoelectric emission only just occurs
  • KEmax depends only on the frequency of the incident photon, and not the intensity of the radiation
  • The majority of photoelectrons will have kinetic energies less than KEmax
29
Q

What would the gradient and y-intercept of a graph of KEmax against frequency be equal to?

A

gradient = h
y-intercept = -Φ

30
Q

What is threshold frequency?

A

The minimum frequency of incident electromagnetic radiation required to remove a photoelectron from the surface of a metal

31
Q

What is threshold wavelength?

A

The longest wavelength of incident electromagnetic radiation that would remove a photoelectron from the surface of a metal

32
Q

What is the work fuction?

A

The minimum energy required to release a photoelectron from the surface of a metal

33
Q

When can an electron escape the surface of a metal?

A

If it absorbs an energy equal to or higher than the value of the work function

34
Q

What is the maximum KE of photoelectrons independent of?

A

The intensity of the incident radiation (KE is only dependant on the frequency of the incident radiation)

35
Q

What is the rate of emission of photoelectrons (photoelectric current) directly proportional to?

A

The intensity of the incident radiation

  • Intensity is proportional to the number of photons striking per second
  • Photoelectrons absorb a single photon only so its also proportional
36
Q

What are electron diffraction tubes used to investigate?

A

The wave properties of electrons

37
Q

How do electron diffraction tubes work?

A
  • The electrons are accelerated in an electron gun to a high potential (such as 5000 V) and are then directed through a film of graphite
  • The electrons diffract from the gaps between carbon atoms and produce a circular pattern on a fluorescent screen made from phosphor
38
Q

In electron diffraction, what does increasing the voltage between the anode and cathode do?

A

Causes the energy, and hence speed, of the electrons to increase. The kinetic energy of the electrons is directly proportional to the voltage across the anode-cathode

KE = 1/2 mv^2 = eV

39
Q

What happens if an electron gun fires at a thin piece of polycrystalline graphite?

A
  • It has carbon atoms arranged in many different layers
  • The electrons pass between the individual atoms in the graphite
  • The gap between the atoms is so small that is it similar to the wavelength of the electrons
  • Therefore, the electrons diffract as waves and form a diffraction pattern seen on the end of the tube
  • The diffraction pattern is observed on the end of the screen as a series of concentric rings
  • It is observed that a larger accelerating voltage reduces the diameter of a given ring and vice versa
40
Q

What are the De Broglie wavelength equations?

A

λ = h / p

λ = h / mv

E = p^2 / 2m. (from KE equation)

λ = h / √ 2mE

λ = de broglie wavelength
p = momentum
m = mass
v = velocity
E = kinetic energy

A-Level Physics (23 decks)

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