Quantum Phenomena

Question 1

What are the 2 requirements for the photoelectric effect to occur?

Answer 1

1) Radiation incident must be above threshold frequency.
2) The energy gained from a single photon must exceed the work function.

Question 2

What varies the threshold frequency?

Answer 2

The type of metal the EM wave is incident on.

Question 3

How does the effect change if you increase intensity?

Answer 3

Rate of emission is proportional to the intensity as only one photon is absorbed by each electron. This also doesn’t affect the Ek of the electrons as they only absorb hf.

Question 4

Explain process of PE effect:

Answer 4

• EM radiation or visible light is incident on the metal.
• Conduction electron absorbs a single photon and gains energy hf.
• If this energy exceeds the work function then it is emitted and any excess energy is for Ek.

Question 5

Equation for kinetic energy of electron emitted:

Answer 5

Ek = hf - work function. the kinetic energy must be greater than 0 for emission to occur.

Question 6

Define stopping potential:

Answer 6

The minimum potential given to the metal plate to stop electron emission. This is because Ek is reduced to zero because of the work done against force of attraction of the metal plate.

Question 7

Explain how emission is quantised:

Answer 7

The energy needed to move up an energy level or escape are multiples of an amount = hf. They need these exact values.

Question 8

What happens to a conduction electron if the energy of the photon incident is not more than the work function:

Answer 8

It absorbs this extra energy an vibrates faster (it already vibrates from its thermal energy) and then quickly loses this energy via collisions with other electrons.

Question 9

Explain how vacuum photocell works:

Answer 9

Light is incident on a photocathode which is in a circuit with an anode opposite it.
The electrons emitted from the cathode are attracted to the anode which is recorded by the microammeter. This value is proportional to the number of electrons emitted per second that land on the anode.

The no. emitted per second is = I/e

Question 10

Explain how work function of electrons emitted can be found using a photocell:

Answer 10

Plot Ek max against frequency. The straight-line of gradient h intercepts the x axis at the threshold frequency and at the y axis the work function.

Question 11

Give equation for electron volts:

Answer 11

work done = qV where q = charge of an electron.

Question 12

What happens if a colliding electron does not have enough energy to cause emission?

Answer 12

It is deflected by the atom and there is no loss in Ek therefore there is no loss in the flow of charge / current.

Question 13

Define excitation energy:

Answer 13

Energy value at which an atom absorbs energy.

Question 14

How do energy levels in an atom work?

Answer 14

An electron in an energy level further away from the nucleus has more energy as it requires it to overcome the electrostatic force of attraction and the potential at that point in the field.

Question 15

Why does gas emit visible light when it conducts electricity?

Answer 15

The atoms absorb energy and excite via collisions with other electrons. When an electron then de-excites, it emits a photon of visible light hence the gas emits light. The excitation only happens when there is a vacancy in the next shell up and when the energy of the photon exactly matches the energy needed to move up a level. However, it does not need to be exact for ionisation, it just needs to be equal to or greater than.

Question 16

Explain how fluorescent tube works: What are the benefits?

Answer 16

Tube that is coated on the inner wall with fluorescent paint and filament electrode at each end. Starter unit required as mains voltage too low to ionise the vapour.
-When the tube is switched on, there is a potential across the ends of the tube and the electrons passing through collide with the mercury atoms which excite and ionise.
-When they de-excite, they emit ultraviolet as photons.
- The ultraviolet photons are absorbed by the mercury atoms causing more excitation.
-Very efficient as little energy is wasted through heat as there is only low power supplied.

Question 17

Explain why line spectra are useful:

Answer 17

Each element has its own line spectra as each atom has distinct energy levels. Therefore, if we know the line spectra we can tell which element it came from.

Question 18

Example of wavelike nature of light and particle like nature of light:

Answer 18

Wave like - diffraction: when wave passes through an aperture and the wave spreads out, like water.
Particle like - photoelectric effect as the electrons absorb a discrete amount of energy which would not happen if it was purely a wave.

Question 19

Evidence of particle like nature of particles and wave like nature of particles

Answer 19

Particle like - Electrons in a beam can be deflected by a magnetic field.
Wave like - electron diffraction

Question 20

Evidence for de Broglie’s hypothesis:

Answer 20

-Electrons fired at metal foil in a vacuum.
-The positive ions in the grains of the ion diffract the electrons.
-Rings on fluorescent screen behind foil is formed due to the fact the electrons are diffracted through the same angle by the grains.
(-the beam of electrons is created by attracting electrons from heated filament wire to a positive metal plate with a hole in the centre. The electrons that pass through the hole form the beam - by increasing the pd the speed is increased so the wavelength decreases which changes the diffraction rings pattern as less diffraction occurs at a different wavelength)

Question 21

What was De Broglie’s hypothesis?

Answer 21

Matter particles have a dual wave-particle nature and these particles have a de Broglie wavelength which is related to its momentum. The wavelength can be changed by changing the velocity of the particle.

Hence wavelength = h/mv