# A2 Advanced Physics I

- The Advanced Physics I exam tests the content on this page

**as well as**the Core Physics I page.

**97** Impulse:

**Impulse**, $\mathrm{\Delta}p$, is the change in momentum, calculated with $F\mathrm{\Delta}t=\mathrm{\Delta}p$ and therefore measured in $\mathrm{N}\text{}\mathrm{s}$ or $\mathrm{k}\mathrm{g}\text{}\mathrm{m}\text{}{\mathrm{s}}^{-1}$- It can also be calculated from the area under a force-time graph
- The formula can be rearranged to $F=\frac{\mathrm{\Delta}p}{\mathrm{\Delta}t}$, showing that force is equal to the rate of change of momentum: Newton's second law of motion

**98** Investigating the relationship between force exerted and change in momentum (CP9):

- Create a surface for a trolley. Prop one end up with some books to compensate for friction. Once it is being accurately compensated for, the trolley should not accelerate or decelerate after being pushed
- Alternatively, use an air track
- Set up two light gates, about $\frac{1}{4}$
^{th}of the total length away from each end - Attach a hanging mass to the trolley with a pulley and string, with several individual masses on
- Now release the trolley from the top of the track. It should be accelerated by the mass and pass through the light gates
- Now find ${m}_{\mathrm{s}\mathrm{y}\mathrm{s}\mathrm{t}\mathrm{e}\mathrm{m}}$. This is the total mass of the system; the sum of the trolley mass, string mass and the hanging mass
- Use the light gates to calculate the change in velocity. Find the change in momentum with $\mathrm{\Delta}{p}_{\mathrm{s}\mathrm{y}\mathrm{s}\mathrm{t}\mathrm{e}\mathrm{m}}={m}_{\mathrm{s}\mathrm{y}\mathrm{s}\mathrm{t}\mathrm{e}\mathrm{m}}\mathrm{\Delta}v$
- Now use the time taken for the trolley to travel between the two light gates, $\mathrm{\Delta}t$, to find the rate of change of momentum with $\frac{\mathrm{\Delta}{p}_{\mathrm{s}\mathrm{y}\mathrm{s}\mathrm{t}\mathrm{e}\mathrm{m}}}{\mathrm{\Delta}t}$
- Now calculate the total force acting on the system with $F=Mg$ where $M$ is the mass hanging from the pulley
- Do three repeats
- Now try varying the hanging mass. Do this by moving masses onto the trolley so that the total system mass remains the same
- Plot force acting on the system against the rate of change of momentum of the system. It should be a straight line. This shows that $F=$ $\frac{\mathrm{\Delta}{p}_{\mathrm{s}\mathrm{y}\mathrm{s}\mathrm{t}\mathrm{e}\mathrm{m}}}{\mathrm{\Delta}t}$. This rearranges to $\mathrm{\Delta}p=F\mathrm{\Delta}t$

**99** Conservation of linear momentum:

- Momentum is a vector so it can be treated separately for each axis
- Therefore, the vertical momentum before and after the collision should be equal. The same applies for horizontal momentum
- An elastic collision between a moving and stationary object (with equal mass) will cause each to move at right angles to each other. This will not work if the collision is completely head-on, but this is extremely unlikely

**100** Analysing collisions between small spheres (CP10):

- Set up a video camera above a table so that it has a view from above. Place metre rulers on the table; one vertical and one horizontal
- Place a stationary sphere (e.g. ball bearing) so that each ruler will read about $50\text{}\mathrm{c}\mathrm{m}$
- Start recording and roll a second sphere towards the stationary one
- Use video analysis software to find the distance travelled each frame. If you know the frame period (i.e. 1/framerate), you can calculate the velocity for each sphere immediately before and immediately after the collision
- The total momentum before and after the collision in each dimension and overall should be equal

**101** Collisions:

- An
**elastic**collision is one where momentum and kinetic energy is conserved (so no energy is dissipated as other forms of energy) - In an
**inelastic**collision, momentum**is**conserved, but kinetic energy**isn't**

**102** Calculating kinetic energy from momentum:

- ${E}_{\mathrm{k}}=$$\frac{{p}^{2}}{2m}$ where $p$ is the momentum and $m$ is the mass of the object
- The formula only works for non-relativistic particles (i.e. particles which have a velocity much less than the speed of light)
- You need to know how to derive this formula:

- Start with $p=mv$ and ${E}_{\mathrm{k}}=\frac{1}{2}m{v}^{2}$

- To substitute the first into the second, square the first: ${p}^{2}={m}^{2}{v}^{2}$

- Now rearrange this to ${v}^{2}=$ $\frac{{p}^{2}}{{m}^{2}}$

- Substitute this into the second equation: ${E}_{\mathrm{k}}=$ $\frac{1}{2}\frac{m{p}^{2}}{{m}^{2}}$

- This simplifies to ${E}_{\mathrm{k}}=$$\frac{{p}^{2}}{2m}$

**103** Radians:

- Angle in radians $=$ $\frac{\pi}{180}$ $\times $ angle in degrees
- Angles measured in radians usually have the symbol 'rad' - e.g. $1.5\text{}\mathrm{r}\mathrm{a}\mathrm{d}$

## Angular displacement:

**Angular displacement**is the angle through which a rotating object has rotated, usually measured in radians

**104** Angular velocity:

**Angular velocity**is the angle an object rotates through per second, measured in $\mathrm{r}\mathrm{a}\mathrm{d}\text{}{\mathrm{s}}^{-1}$ with the symbol $\omega $- $\omega =\frac{\theta}{t}$ where $\theta $ the angular displacement (radians), and $t$ is that time period in seconds
**Linear velocity**is the actual velocity of a point if a tangent were drawn from it. It has the symbol**$v$**- $v=r\omega $ where $r$ is the distance from the centre of the circle to the point which you're calculating the velocity of

## Period:

- The
**period**, $T$ is the time taken for a full revolution (in seconds) - If you know the frequency (revolutions per second), you can use the equation $T=\frac{1}{f}$
- The following can also be derived: $T=$ $\frac{2\pi}{\omega}$

**105** Centripetal acceleration:

- If an object is moving in a circle at constant speed, it is still
*accelerating*because the direction component of its velocity is changing. This acceleration is directed towards the centre of the circle and called**centripetal acceleration**,**$a$** - So centripetal acceleration is the rate of change of linear velocity
- $a=$ $\frac{{v}^{2}}{r}$ and $a$ = $r{\omega}^{2}$

## Derivation:

- Let $u$ be the initial linear velocity and $v$ be the linear velocity after a movement of $\theta $ radians. Let $x$ be the distance between the reading of $u$ and the reading of $v$ around the circle (shown below):
- We can now create a vector triangle containing $u$, $v$ and $\mathrm{\Delta}v$:
- This is a sector, so we can use the sector equation for arc length $s=\theta r$ to get $\theta =$ $\frac{x}{r}$
- When $\theta $ is very small, the change in velocity $\mathrm{\Delta}v$ will be approximately equal to the arc length $x$. This can be written as $\theta v=\mathrm{\Delta}v$
- If we combine these two equations, we get $\mathrm{\Delta}v=$ $\frac{x}{r}$ $\times \text{}v$
- When $\theta $ is very small, $x=vt$. Therefore, this can be substituted into the above equation to get $\mathrm{\Delta}v=$ $\frac{{v}^{2}t}{r}$
- If we divide both sides by $t$, we get $a=$ $\frac{{v}^{2}}{r}$
- To get $a=r{\omega}^{2}$, substitute $v=\omega r$ into this formula

**106** Centripetal force:

- If an object is accelerating, it must have a force acting on it (Newton's first law)
**Centripetal force**acts perpendicular to an object's linear velocity, towards the centre of the circle- If centripetal force suddenly stopped, the object would fly off at a tangent to its motion when the force stopped

**107** Centripetal force:

- You can use the below equations to do centripetal force calculations
- $F=ma=$ $\frac{m{v}^{2}}{r}$ $=mr{\omega}^{2}$

**108** Electric fields:

- An
**electric field**is a region where a charged particle experiences a force

**109** Electric field strength:

**Electric field strength**, $E$, is the force per unit of positive charge- It is a vector in the direction that a positive charge would move
- If a charge $Q$ experiences a force $F$, $E=$ $\frac{F}{Q}$
- From this, we can see that the units for $E$ are $\mathrm{N}\text{}{\mathrm{C}}^{-1}$

**110** Coulomb's law:

- If two charges are further from each other, the force is lower. This follows an inverse square law:

$F\propto $ $\frac{1}{{r}^{2}}$, where $r$ is the distance between the charges in $\mathrm{m}$ - Force is also proportional to charge ($F\propto Q$). The two equations can be combined to work with two charges:

$F=$ $\frac{{Q}_{1}{Q}_{2}}{4\pi {\epsilon}_{0}{r}^{2}}$ - $F$ is the force of one of the charges. The force on the other is equal but opposite (i.e. $-F$)
- If the force is attractive, $F<0$. If it is repulsive, $F>0$
- ${\epsilon}_{0}$ is a constant depending on the substance. In air or a vacuum, it's the
**permittivity of free space**: ${\epsilon}_{0}=8.85\times {10}^{-12}$ $\mathrm{F}\text{}{\mathrm{m}}^{-1}$

**111** Calculating the electric field strength due to a point charge:

- At a distance $r$ $\mathrm{m}$ from a point charge $Q$, $E=$ $\frac{Q}{4\pi {\epsilon}_{0}{r}^{2}}$

**112** Electric potential:

**Electric potential**at a point in an electric field is the work done per unit charge to move a positively charged particle from infinity to that point- For example:

- if you have a negative point charge, call this**C**

- pick a random point on its infinitely large electric field, and call this point**P**

- the electric potential at point**P**is the energy which would be needed to attract a positive charge from infinity distance to**P**

**113** Parallel plates:

- With parallel plates with a potential difference between them, there will be an electric field
- This field is constant at all positions between the two plates
- Where $V$ is the potential difference between the plates and $d$ is the distance between them, $E=$ $\frac{V}{d}$

**114** Electric potential at a point in a point charge electric field:

- Electric potential (the energy to move a $+1\text{}\mathrm{C}$ charge from infinity to a point) is calculated with the equation $V=$ $\frac{Q}{4\pi {\epsilon}_{0}r}$ where $Q$ is the charge of the point charge and $r$ is the distance between the point charge and $+1\text{}\mathrm{C}$ charge

**115** Electric field lines:

**Electric field lines**show the direction that a positive charge would take in an electric field- The spacing between the lines show the strength of the field - stronger fields have field lines drawn closer together

## For parallel plates:

- Parallel plates have a
**uniform electric field**because the charge is uniform on all points between the two plates - Therefore, the electric field lines are drawn like this (from positive to negative):

## For point charges:

- Point charges have a
**non-uniform electric field**- the charge on a positive charge would be higher if it is closer to the point charge. On the diagram, the lines are therefore closer together nearer to the point charge:

## Lines of equipotential:

**Lines of equipotential**join up points at the same potential- They should be drawn at uniform intervals - e.g. with parallel plates with a potential difference of $100\text{}\mathrm{V}$, you could draw an equipotential line every $25\text{}\mathrm{V}$
- With parallel plates, the lines of equipotential are straight and perpendicular to the field lines. They have an equal distance between them
- With point charges, the lines of equipotential are circular. They are closer together closer to the point charge

**116** Capacitors:

- A capacitor is a component that can store electrical charge
- They contain two conducting metal plates, separated by an insulating material or gap

## Capacitance:

- The amount of charge a capacitor can store is its
**capacitance**, $C$, measured in**farads** - $C=$ $\frac{Q}{V}$ where $Q$ is the charge and $V$ is the pd

**117** Deriving $W=\xbd\text{}\text{}\text{}QV$:

- The potential difference across a capacitor is directly proportional to the charge stored in it
- Therefore, a graph of potential difference against charge stored would be a straight line through the origin
- Because in a triangle, area $=\frac{1}{2}\times $ base $\times $ height, the energy stored by a capacitor can be calculated with $W=\frac{1}{2}QV$

## Derivations from $W=\xbd\text{}\text{}\text{}QV$:

- By combining the equations $Q$ $=CV$ and $W=\frac{1}{2}$$Q$$V$, we get $W=\frac{1}{2}CVV=\frac{1}{2}C{V}^{2}$
- By combining the equations $V$ $=\frac{Q}{C}$ (rearranged from $Q=CV$) and $W=\frac{1}{2}C$$V$${}^{2}$, we get $W=$ $\frac{1}{2}\frac{C{Q}^{2}}{{C}^{2}}$ $=$ $\frac{\frac{1}{2}{Q}^{2}}{C}$
- So with these two methods, we have two further equations:

$W=\frac{1}{2}C{V}^{2}$ and $W=$ $\frac{\frac{1}{2}{Q}^{2}}{C}$

*(the colour coding is slightly different here for clarity)*

**118** Resistor-capacitor charging curves:

- In a circuit with a resistor and capacitor, charging/discharging curves are exponential functions if drawn on a graph of pd/charge/current against time
- Because these graphs are powers of $e$, the first half-life will equal the second half-life and so on. This will also work for any fraction - e.g. quarter-lives. This is due to the
*constant ratio*property of exponential relationships - The
**time constant**, $RC$ is the time taken for a capacitor's pd/charge/current to fall to $\frac{1}{e}$ (about $37\mathrm{\%}$) of its original value. So after two time constants, it will have fallen to $0.37\times 0.37\approx 14\mathrm{\%}$ of its original pd/charge/current - As implied by $RC$, it is calculated by multiplying the capacitor's capacitance and the resistance of the resistor (assuming that the rest of the circuit has negligible resistance)
- After five time constants of discharging, the charge in the capacitor has fallen below $1\mathrm{\%}$ of its original charge, which can be considered zero coulombs in most cases

**119** Analysing capacitor charging and discharging (CP11):

- If you connect a
**cathode ray oscilloscope**(CRO) in parallel across a capacitor, you can charge it and discharge it through a resistor and view the potential difference on the display - By setting the
**timebase**(number of milliseconds per division on the horizontal axis) to 0, there'll be a dot in the centre. This will show the potential difference - Set the
**gain**(number of volts per division on the vertical axis) so that the peak pd will be at the top - Now use a timer to record the amount of time for it to fall/rise by one division. Do this for several divisions. This can be used to plot a $V$-$t$ graph of charging or discharging
- Instead of a CRO, you could use a data logger. This will connect to an ammeter (in series) and voltage sensor (in parallel) and plug into a computer

**120** Capacitor charge equations:

- $Q={Q}_{0}{e}^{-t/RC}$ where

- $Q$ is the charge at the time $t$

- ${Q}_{0}$ is the initial charge when $t=0$

- $t$ is the time after the capacitor started charging/discharging that you are investigating

- $R$ is the resistance of the circuit

- $C$ is the capacitance of the capacitor - $\mathrm{\Delta}Q=I\times \mathrm{\Delta}t$ can be substituted into this equation to derive $I={I}_{0}{e}^{-t/RC}$
- The above equation, along with $V=IR$ are used to derive $V={V}_{0}{e}^{-t/RC}$
- By taking natural logs of both sides, $\mathrm{ln}Q=\mathrm{ln}{Q}_{0}-\frac{t}{RC}$ can be derived (and the same for $I$ and $V$). When deriving this, remember that the log of ${Q}_{0}{e}^{-t/RC}$ is $\mathrm{ln}({Q}_{0}\times {e}^{\frac{-t}{RC}})$ and
**not**$\mathrm{ln}({Q}_{0})\times \mathrm{ln}({e}^{\frac{-t}{RC}})$

**121** Magnetic flux:

**Magnetic flux density**, $B$ is a measure of the strength of a magnetic field, measured in $\mathrm{T}$ (Teslas)**Magnetic flux**, $\mathrm{\Phi}$, measures the total magnetic field passing through a given area. It's highest if the field lines are perpendicular to this area (they pass through at a 90° angle). Magnetic flux is calculated with $\mathrm{\Phi}=BA\mathrm{cos}\theta $ [not given in the exam]. $\theta $ is the angle between the field lines and**normal**of the area. Magnetic flux is measured in Webers ($\mathrm{W}\mathrm{b}$)**Magnetic flux linkage**, $\mathrm{\Phi}N$ is the flux of a coil, $\mathrm{\Phi}$, multiplied by the number of turns, $N$

**122** Fleming's left-hand rule for motors:

- Thu
**M**b:**M**otion **F**irst finger:**F**ield (N to S)**C**entre finger (middle finger):**C**urrent ($+$ to $-$)

## Charged particles moving through a magnetic field:

- $F=Bqv\mathrm{sin}\theta $ where

- $F$ is the force acting on the particle, in $\mathrm{N}$

- $q$ is the charge of the particle, in $\mathrm{C}$

- $v$ is the velocity of the charged particle, in $\mathrm{m}\text{}{\mathrm{s}}^{-1}$

- $\theta $ is the angle between the particle's direction and the field lines - Using Fleming's left-hand rule, you can see that a charged particle will be deflected in a circular path when through a magnetic field

**123** Current carrying conductors in a magnetic field:

- $F=BIl\mathrm{sin}\theta $ where

- $F$ is the force experienced by a wire, in $\mathrm{N}$

- $B$ is the magnetic flux density of the field, in $\mathrm{T}$

- $I$ is the current travelling through the wire, in $\mathrm{A}$

- $l$ is the length of wire, in $\mathrm{m}$

- $\theta $ is the angle between the current direction and field lines

**124** Inducing emf in coils:

- An emf is generated whenever a conductor cuts field lines (or when the flux linkage changes)
- This emf can be increased by

- increasing magnetic field strength

- having more turns of the coil

- increasing the rate of field cutting

**125** Mutual inductance:

- When current passes through a coil, a magnetic field is produced. If this is near to another coil, an emf will be induced in the second coil whenever the magnetic field of the first changes
- Therefore, an emf can be induced in a second coil by passing an alternating current through the first
- This property is used in items such as wireless phone chargers (wirelessly transferring power over short distances) and transformers (increasing or decreasing the voltage of AC)
- The induced emf can be increased by

- connecting the coils with a permanent magnet (e.g. iron rod)

- decreasing the distance between the two coils

- increasing the AC power input

**126** Lenz's law:

- Lenz's law says that the emf induced is induced in the direction opposite to the change that caused it
- If you consider the direction of the induced emf to be the direction of motion, you can apply Fleming's left-hand rule

**127** Faraday's law:

- $\epsilon =$ $\frac{-\mathrm{d}(N\mathrm{\Phi})}{\mathrm{d}t}$, where

- $\epsilon $ is the induced emf ($\mathrm{V}$)

- $\mathrm{d}(N\mathrm{\Phi})$ is the change in flux linkage ($\mathrm{W}\mathrm{b}$)

- $\mathrm{d}t$ is the change in time ($\mathrm{s}$) - This equation shows that the induced emf is proportional to the rate at which the flux is cut
- The fraction is negative due to Lenz's law

**128** Alternating current:

- The
**period**of an AC is the time taken for one**complete cycle** **Frequency**is the number of complete cycles per second- The
**peak value**of an AC is the maximum positive/negative value

## Root-mean-square value:

- The
**root-mean-square value**, or**rms**, is a way of measuring the current/voltage of an AC (since they'll vary depending on the current stage in the cycle) - The
**rms voltage**is the voltage that a DC would need to be at to produce the same power output - The rms voltage is lower than the peak voltage because an alternating current spends some time at or near to $0\text{}\mathrm{V}$ in each cycle
- The rms current is written as ${I}_{\mathrm{r}\mathrm{m}\mathrm{s}}$ and rms voltage is written as ${V}_{\mathrm{r}\mathrm{m}\mathrm{s}}$

**129** Converting between peak and rms values:

- Peak voltage/current and rms voltage/current are related with the following equations:

${V}_{\mathrm{r}\mathrm{m}\mathrm{s}}=$ $\frac{{V}_{0}}{\sqrt{2}}$ and ${I}_{\mathrm{r}\mathrm{m}\mathrm{s}}=$ $\frac{{I}_{0}}{\sqrt{2}}$

**130** Proton/nucleon numbers:

- The
**proton number**(or**atomic number**), $Z$, is the number of protons in the nucleus of an atom/ion. It is equal to the number of electrons if the atom is neutral - The
**nucleon number**(or**mass number**), $A$, is the number of protons and neutrons in the nucleus

**131** Understanding of atomic structure:

**Dalton**: originally suggested that everything is made up of small particles (atoms), each with the shape of a simple sphere**Thomson**: discovered the electron, creating the 'plum pudding' model - negatively charged electrons embedded inside a positively charged sphere**Rutherford/Geiger/Marsden**: fired alpha particles (positive helium nuclei made up of 2 neutrons and 2 protons) through thin gold foil. They noticed that most passed straight through, but some were refracted (due to proton-proton repulsion) and some were reflected (when they were repelled by the nucleus due to being very close). This caused the model to change to the nuclear model, a positive nucleus surrounded by negative electrons

## Conclusions from the gold foil experiment:

- Most alpha particles passed straight through,
**so the atom is mostly empty space** - Some alpha particles were deflected,
**so the centre of the atom must have a positive charge to repel them** - Very few particles were deflected through angles greater than 90°,
**so the nucleus must be very small** - Alpha particles would only be deflected greater than 90° if they were scattered by something with a higher mass than themselves,
**so the nucleus must make up most of the mass of the atom**

**132** Thermionic emission:

- When a metal is heated, its free electrons gain energy
- With enough energy, they'll overcome the attraction from the positive metal ions and leave the metal. This is called
**thermionic emission** **Electron guns**produce a thin beam of high-speed electrons

- A high voltage is applied to a heating coil

- This coil heats a cathode ($-$) (which will have free electrons)

- Electrons are emitted

- A short distance from the cathode is a cylindrical anode ($+$) with a small hole

- This anode attracts the electrons, accelerating them. Those travelling in a straight line will pass through the hole- Magnetic fields are also sometimes used to change the course or accelerate electrons

**133** Linacs:

- A
**linear particle accelerator**(*linac*) is a linear tube lined with electrodes - Each electrode has the opposite charge to the adjacent ones
- Charged particles are fired from the start of the tube
- These electrodes are attached to an alternating current, at a frequency so that the particles are always being repelled from the electrode they just left and attracted to the one they are approaching
- This accelerates the particles
- Each electrode is longer than the last so that the particles spend the same amount of time in each

## Cyclotrons:

**Cyclotrons**accelerate particles in a spiral shape before releasing them- They contain two thin hollow semicircular electrodes with an alternating (+ and -) potential difference so that the particles are always accelerating
- There is also a magnetic field perpendicular to the electrodes
- Charged particles are fired into an electrode. Due to the magnetic field, it follows a circular path
- The radius slowly increases as the particle accelerates, until it reaches the edge and leaves the cyclotron

## Detectors - ionisation:

- Detectors used in particle physics experiments can contain a gas
- This gas is ionised by charged particles
- The gas ions are attracted to wires with a current flowing. The current spike caused by these ions can be detected

## Detectors - deflection:

- A strong magnetic field is applied across the detector
- Colliding particles will travel in the same direction so they will not be affected
- Particles produced in collisions will be deflected by the magnetic field, and the deflection can be used to calculate their mass and velocity

**134** Finding the radius of a charged particle in a magnetic field:

- $r=$ $\frac{p}{BQ}$. This is derived as follows:
- The centripetal force equation is $F=$ $\frac{m{v}^{2}}{r}$
- We also have $F=BQv$
- These are both equal to each other, so $\frac{m{v}^{2}}{r}$ $=BQv$
- Multiplying both sides by $r$: $m{v}^{2}=BQvr$
- Dividing both sides by $v$: $mv=BQr$
- Dividing both sides by $BQ$: $r=$ $\frac{mv}{BQ}$ $=$ $\frac{p}{BQ}$

**135** Interpreting particle tracks:

- Charge, momentum and energy are always maintained in a particle collision
- When a charged particle passes through a substance, it ionises the atoms is passes through. This leaves a trail of ions which can be detected to find the path of electrons
**Cloud chambers**contain a supercooled vapour. Ions condense the vapour, leaving a track. More ionising particles will leave short, thick tracks**Bubble chambers**contain hydrogen under pressure to keep it above its boiling point as a liquid. Ions cause gas bubbles to form- Neutral particles won't show up on a particle track. If they then split into charged particles, those will be visible. This is what causes V shapes - two oppositely charged particles leaving from a decayed neutral particle

**136** Why high energies are required in particle physics experiments:

- To get a positively charged particle close to a nucleus, it must have enough energy to overcome the forces of repulsion
- The higher the collision energy, the larger the produced particles will be (due to mass-energy equivalence)
- The more energy that can be given to particles, the shorter their wavelength and the higher the resolution if using them as a probe (e.g. an electron microscope has higher resolution than a light microscope because electrons have a smaller wavelength. The same applies here)

**137** Mass-energy equivalence:

- Mass and energy are related, shown in the equation $\mathrm{\Delta}E={c}^{2}\mathrm{\Delta}m$, where

- $c$ is the speed of light in a vacuum

- $\mathrm{\Delta}E$ is the energy of the particle in $\mathrm{J}$

- $\mathrm{\Delta}m$ is the mass in $\mathrm{k}\mathrm{g}$

**138** Writing particle masses:

- The equation $\mathrm{\Delta}E={c}^{2}\mathrm{\Delta}m$ can be rearranged to $\mathrm{\Delta}m=$ $\frac{\mathrm{\Delta}E}{{c}^{2}}$
- Therefore, the mass of a particle can be expressed in terms of its energy (e.g. the mass of an electron at rest is about $9.11\times {10}^{-31}$ $\mathrm{k}\mathrm{g}$ or 0.512 $\mathrm{M}\mathrm{e}\mathrm{V}/{c}^{2}$)
- To convert to $\mathrm{e}\mathrm{V}/{c}^{2}$ form from a mass:

- First, multiply the mass in $\mathrm{k}\mathrm{g}$ by ${c}^{2}$, $(3.00\times {10}^{8}{)}^{2}$

- Then divide by the charge of an electron, $1.60\times {10}^{-19}$ $\mathrm{C}$

- Round to an appropriate number of s.f. and use SI prefixes (e.g.**M**ega or**T**era) to make the number more readable

- e.g. 512 438 $\mathrm{e}\mathrm{V}/{c}^{2}$ → 0.512 $\mathrm{M}\mathrm{e}\mathrm{V}/{c}^{2}$

**139** Relativistic particles:

- A
**relativistic**speed is a speed close to the speed of light - At these speeds, particles experience
**time dilation**- they experience time more slowly - Therefore, they will last longer, and decay more slowly
- Many equations, such as ${E}_{\mathrm{k}}=\frac{1}{2}m{v}^{2}$, are not accurate for relativistic particles

**140, 141** Baryons (e.g. neutrons and protons):

- There are 6 types of
**quark**; 12 in total including their anti-quarks

- Up, charm and top quarks have charge $+2e/3$, in ascending mass

- Down, strange and bottom quarks have charge $-1e/3$, in ascending mass - Quarks are always part of another particle, they cannot exist on their own
- A
**baryon**is a particle made of three quarks, such as a proton ($uud$) and neutron ($udd$) **Anti-quarks**are written with a line above, e.g. $\overline{u}$ and $\overline{d}$

## Antimatter:

- Every fundamental particle has an antimatter version with identical mass and an identical strength, but opposite charge
- Anti-protons are written ${p}^{-}$ and anti-electrons (
*positrons*) are written ${e}^{+}$. Otherwise, anti-particles are written with a bar above - This symmetry is how the top quark was predicted
- Anti-atoms also exist - for example, anti-hydrogen is made of an anti-proton and positron

## Mesons:

**Mesons**are particles made from a quark and an anti-quark**Pions**are the most common mesons and are written ${\pi}^{n}$ where $n$ is the charge (either $-$, $+$ or 0)

- e.g. an up quark and an anti-up quark is called "pi zero" or ${\pi}^{0}$

## Leptons:

**Leptons**, e.g. electrons ($e$) and positrons (${e}^{-}$), are fundamental particles- Leptons have $-1e$ charge [and their anti-leptons are $+1e$, indicated with a $+$ or $-$ above them (e.g. ${\mu}^{+}$ is an anti-muon)]

## Photons:

**Photons**are packets or energy with no mass and no charge

**142** Particle interactions:

- On both sides of a particle equation, certain quantities need to be conserved:

-**Baryon number**, $B$ (quarks have $B=\frac{1}{3}$, anti-quarks have $B=-\frac{1}{3}$)

-**Lepton number**, ${L}_{n}$ where $n$ is the generation ($e$, $\mu $ or $\tau $). Leptons have $L=1$, anti-leptons have $L=-1$

-**Charge**, $Q$ - Mass and energy
**don't**need to be conserved (due to $\mathrm{\Delta}E={c}^{2}\mathrm{\Delta}m$)