Quantum Physics

Question 1

How does electromagnetic energy exsist?

Answer 1

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

Question 2

What is the photon model?

Answer 2

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

Question 3

What are the packets that EM radiation come in called?

Answer 3

Photons

Question 4

What are the equations for the energy of a photon?

Answer 4

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

Question 5

What is the value of Plancks constant?

Answer 5

6.63 x 10^-34 Js

Question 6

At subatomic level, what unit for energy is used?

Answer 6

The electronvolt

Question 7

What is 1 electron volt equal to?

Answer 7

1.6 x 10^-19 J

Question 8

How do you convert from electron volts to joules?

Answer 8

multiply by 1.6 x 10^19

Question 9

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

Answer 9

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

Question 10

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

Answer 10

The energy of a single photon

Question 11

What is the equation for energy lost by each electron?

Answer 11

E = eV

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

Question 12

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

Answer 12

eV = hc/λ

Question 13

What equation can be used to estimate Plancks constant?

Answer 13

eV = hc/λ

Question 14

How do you determine Plancks constant using LEDs?

Answer 14

Question 15

What is the photoelectric effect?

Answer 15

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

Question 16

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

Answer 16

Photoelectrons

Question 17

What does the photoelectric effect provide evidence for?

Answer 17

The particle nature of light

Question 18

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

Answer 18

• 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

Question 19

What are the 5 main observations of the photoelectric effect?

Answer 19

• 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

Question 20

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’?

Answer 20

• 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

Question 21

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

Answer 21

• 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

Question 22

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

Answer 22

• 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

Question 23

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

Answer 23

• 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

Question 24

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

Answer 24

• 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

Question 25

Explain the one to one interaction of photons and surface electrons

Answer 25

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

Question 26

What is the energy of a single photon equal to?

Answer 26

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

Question 27

What is the work function and kinetic energy equation?

Answer 27

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

Question 28

What does the work function and KEmax equation show?

Answer 28

- 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

Question 29

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

Answer 29

gradient = h y-intercept = -Φ

Question 30

What is threshold frequency?

Answer 30

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

Question 31

What is threshold wavelength?

Answer 31

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

Question 32

What is the work fuction?

Answer 32

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

Question 33

When can an electron escape the surface of a metal?

Answer 33

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

Question 34

What is the maximum KE of photoelectrons independent of?

Answer 34

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

Question 35

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

Answer 35

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

Question 36

What are electron diffraction tubes used to investigate?

Answer 36

The wave properties of electrons

Question 37

How do electron diffraction tubes work?

Answer 37

- 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

Question 38

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

Answer 38

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

Question 39

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

Answer 39

- 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

Question 40

What are the De Broglie wavelength equations?

Answer 40

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