Chapter 3

Question 1

Threshold frequency

Answer 1

The minimum frequency of light that can cause the photoelectric effect

Question 2

The atom

Answer 2

The atom is formed of 3 constituents protons,neutrons and electrons.

Question 3

Specific charge

Answer 3

The SP of a particle is the charge - mass ratio and is calculated by dividing a particles charge by its mass

Charge / mass

Question 4

The photo electric affect

Answer 4

Is where photoelectrons are emitted for the surface of a metal after light above a certain frequency is shone on it. This certain frequency is different for different types of metals and is called the threshold frequency.

Question 5

The threshold frequency

Answer 5

Couldn’t be explained by wave theory, as it suggest that any frequency of light should be able to cause the photoelectric emission as the energy absorbed by each electron will gradually increase with each incoming wave.

Question 6

Threshold frequency explained with the photon model of light

Answer 6

Em waves travel in discrete packets called photons, which have an energy which is directly proportional to the frequency

Each electron can absorb a single photon therefore a photon electron is only emitted if the frequency is above the threshold frequency.

If the intensity of the light is increased if the frequency is above the threshold more photon electrons are emitted per second.

Question 7

What is work function

Answer 7

Work function of a metal is the minimum energy required for electrons to be emitted from the surface of a metal and it is denoted by (circle with a line going straight down)

Question 8

Stopping potential

Answer 8

Stopping potential is the potential difference you would need to apply across the metal to stop the photo electrons with the maximum kinetic energy. Measuring stopping potential allows you to find the maximum kinetic energy of the released photo electrons as EK (max) = eVs. Where Vs is the stopping potential and e is the charge of an electron.

This is derived using the fact that energy = charge x voltage.

Question 9

PE equation

Answer 9

E = h x f = 0| + Ek (max)

Shows relationships between work function, maximum kinetic energy and the frequency of light.

Question 10

Collisions of electrons and excitation

Answer 10

Electrons in atoms can only exist in discrete energy levels, these electrons can gain energy from Collisions with free electrons, which can cause them to move up in Energy levels this is known as excitation

They gain enough energy to be removed from the atom entirely this is called ionisation

Question 11

If an electron becomes excited …

Answer 11

It will quickly return to its original energy level ( ground state) and therefore releases energy it gained in the form of the photon.

Question 12

Fluorescent tube

Answer 12

An example of a practical use of excitation is in a fluorescent tube in order to produce light.
Fluorescent tubes are filled with mercury vapour, across which a high voltage is applied.

This voltage accelerates free electrons through the tube, which collide with the mercury
atoms causing them to become ionised, releasing more free electrons.
The free electrons collide with the mercury atoms, causing them to become excited. When they de-excite they release photons, most of which are in the UV range.
The (phosphorous) fluorescent coating on the inside of the tube, absorbs these UV photons and therefore electrons in the atoms of the coating become excited and de-excite releasing photons of visible light.

Question 13

describing the energy difference between energy levels,

Answer 13

When describing the energy difference between energy levels, the values of energy are very small, therefore the unit, electron volts (eV) is used instead of joules (J).
An electron volt is defined as the energy gained by one electron when passing through a potential difference of 1 volt.

Question 14

Energy levels and photon emission

Answer 14

By passing the light from a fluorescent tube through a diffraction grating or prism, you get a line spectrum.
Each line in the spectrum will represent a different wavelength of light emitted by the tube.
As this spectrum is not continuous but rather contains only discrete
values of wavelength, the only photon energies emitted will correspond to these
wavelengths, therefore this is evidence to show that electrons in atoms can only transition between discrete energy levels,

Question 15

line absorption spectrum

Answer 15

which looks like a continuous spectrum of all possible wavelengths of light, with black lines at certain wavelengths.
These lines represent the possible differences in energy levels as the atoms in the gas can only absorb photons of an energy equal to the exact difference between two energy levels.

Question 16

Difference between two energy

Answer 16

difference between two energy levels is equal to a specita photon energy emitted by a
nuorescent tube, or absorbed in a line absorption spectrum. Therefore: ΔE = E1 - E2
where E1 and E2

Question 17

Wave-particle duality

Answer 17

Light can be shown as having both wave and particle properties.
Examples of light acting as a wave are diffraction and interference, while an example of light acting as a particle is the photoelectric effect.

Electrons can also be shown as having both wave and particle properties, as the wave
nature of electrons can be observed through electron diffraction, as only waves can
experience diffraction.

Question 18

De brogile

Answer 18

De Broglie hypothesised that if light was shown to have particle properties, then particles
should also have wave-like properties, and he wrote an equation relating the wavelength (λ)
of an object to its momentum (mv):
入= h/m x V
where h is the planck constant.
Using the above equation you can see how the amount of diffraction changes as a particle’S momentum changes.
When the momentum is increased, the wavelength will decrease, and therefore the amount of diffraction decreases, so the concentric rings of the interference pattern become closer.
Whereas, when momentum is decreased, the wavelength increases, the amount of diffraction increases so the rings move further apart.

Question 19

Ionisation

Answer 19

An ion is a charged atom.
The number of electrons in an ion Is not equal to the number of protons, An ion is formed from an uncharged atom by adding or removing electrons from the atom. Adding electrons makes the atom into a negative ion.
Removing electrons makes the atom into a positive ion.
Any process of creating ions is called jonisation. For example:

Alpha, beta, and gamma radiation create ions when they pass
through substances and collide with the atoms of the substance.

Electrons passing through a fluorescent tube create ions when th
collide with the atoms of the gas or vapour in the tube.

Question 20

De excitation

Answer 20

Process in which an atom loses energy by photon emission as a result of an electron inside an atom moving from an outer shell to an inner shell.

Question 21

De broglies hypothesis

Answer 21

A narrow beam of electrons in a vacuum tube is directed at a thin
Metal foil.
A metal is composed of many tiny crystalline regions.
Each region, or grain, Consists of positive ions arranged in fixed
positions in rows in a regular pattern.
The rows ot atoms cause the electrons in the beam to be dilfracted, just as a beam of light is diffracted when it passes through a slit.

The electrons in the bear pass through the metal toil and are
dittracted in certain directions only.
They form a pattern of rings on a fluorescent screen at the end of the tube.
Each ring is due to electrons diffracted by the same amount from grains of
different orientations, at the same angle to the incident beam.

The beam of electrons is produced by attracting electrons rom a
heated filament wire to a positively charged metal plate, which
has a small hole at its centre.
Electrons that pass through the hole form the beam.
The speed of these electrons can be increased by increasing the potential difference between the filament and the metal plate.
This makes the diffraction rings smaller, because the
increase of speed makes the de Broglie wavelength smaller.
So less diffraction occurs and the rings become smaller.

Question 22

Wave speed

Answer 22

Freq x wavelength

Question 23

Freq =

Answer 23

Wave speed / wavelength

Question 24

Wavelength

Answer 24

Wave speed / frequency

Question 25

Maximum kinetic energy of an emitted electron

Answer 25

Ek max = h x f - 0|

Question 26

E =

Answer 26

H x f

F = E / h

Question 27

Momentum

Answer 27

P = m x v

M = particle mass
P = momentum 
V = speed

Question 28

Speed of alpha particle

Answer 28

1/2 x m x v ² = Ek

Question 29

Plot 1/λ against

Answer 29

Y = m x + c 
Ek = 1/λ

Y  = m   x    +   C 
Ek= hc 1/λ  -   0/

H x f = 0/ + Ek
H x c 1/λ = 0/ + Ek
F = c / λ
Ek = h x c x 1/λ - 0/

Question 30

Ionisation

Answer 30

The process of making atoms into ions. This caused by Energy being transferred to the electron allowing it to escape the atom.

Question 31

Excitation

Answer 31

The process in which an atom absorbs energy, without becoming ionised as a result of an electron inside an atom moving from an inner shell to an outer shell