Quantum Physics Flashcards
How does electromagnetic energy exsist?
Only in certain values - it appeared to come in little packets/ quantas
What is the photon model?
EM radiation comes in tiny packets of energy, rather than one continuous wave
What are the packets that EM radiation come in called?
Photons
What are the equations for the energy of a photon?
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
What is the value of Plancks constant?
6.63 x 10^-34 Js
At subatomic level, what unit for energy is used?
The electronvolt
What is 1 electron volt equal to?
1.6 x 10^-19 J
How do you convert from electron volts to joules?
multiply by 1.6 x 10^19
What happens when a large enough potential difference is applied across an LED?
It emits photons that all the have the same wavelength and frequency.
When an LED just begins to glow, what is the energy lost by each electron converted to?
The energy of a single photon
What is the equation for energy lost by each electron?
E = eV
E = energy lost by electron
e = elementary charge
V = potential difference against the LED
What is the equation for energy lost by electron = energy of photon?
eV = hc/λ
What equation can be used to estimate Plancks constant?
eV = hc/λ
How do you determine Plancks constant using LEDs?
What is the photoelectric effect?
The phenomena in which electrons are emitted from the surface of a metal upon the absorption of electromagnetic radiation
What are the electrons emitted during the photoelectric effect known as?
Photoelectrons
What does the photoelectric effect provide evidence for?
The particle nature of light
How is the photoelectric effect shown with the gold leaf electroscope?
- 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
What are the 5 main observations of the photoelectric effect?
- 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
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’?
- 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
What is the explanation of the observation ‘Using a higher frequency light source does not change how quickly the gold leaf falls ‘?
- 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
What is the explanation of the observation ‘Using a filament light source causes no charge in the gold leafs position’ ?
- 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
What is the explanation of the observation ‘Using a positively charged plate causes no change in the gold leafs position’ ?
- 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
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 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
Explain the one to one interaction of photons and surface electrons
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
What is the energy of a single photon equal to?
Minimum energy required to free a single electron from the metal surface + maximum kinetic energy of the emitted electron
What is the work function and kinetic energy equation?
hf = Φ + KEmax
h = Plancks constant
f = frequency Hz
Φ = work function J
KEmax = maximum kinetic energy
What does the work function and KEmax equation show?
- 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
What would the gradient and y-intercept of a graph of KEmax against frequency be equal to?
gradient = h
y-intercept = -Φ
What is threshold frequency?
The minimum frequency of incident electromagnetic radiation required to remove a photoelectron from the surface of a metal
What is threshold wavelength?
The longest wavelength of incident electromagnetic radiation that would remove a photoelectron from the surface of a metal
What is the work fuction?
The minimum energy required to release a photoelectron from the surface of a metal
When can an electron escape the surface of a metal?
If it absorbs an energy equal to or higher than the value of the work function
What is the maximum KE of photoelectrons independent of?
The intensity of the incident radiation (KE is only dependant on the frequency of the incident radiation)
What is the rate of emission of photoelectrons (photoelectric current) directly proportional to?
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
What are electron diffraction tubes used to investigate?
The wave properties of electrons
How do electron diffraction tubes work?
- 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
In electron diffraction, what does increasing the voltage between the anode and cathode do?
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
What happens if an electron gun fires at a thin piece of polycrystalline graphite?
- 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
What are the De Broglie wavelength equations?
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
