A Level Advanced Physics II

The Advanced Physics II exam tests the content on this page as well as the Core Physics II page.
All derivations on this page need to be memorised (I wouldn't include them if they didn't).

144 Specific heat capacity:

• The specific heat capacity, $c$, of a substance, is the energy required to heat $1\mathrm{k}\mathrm{g}$ of it by $1\mathrm{K}$
• $\mathrm{\Delta }E=mc\mathrm{\Delta }\theta$ where
  - $\mathrm{\Delta }E$ is the energy change in $\mathrm{J}$
  - $m$ is the mass in $\mathrm{k}\mathrm{g}$
  - $c$ is the specific heat capacity in $\mathrm{J}{\mathrm{k}\mathrm{g}}^{-1}{\mathrm{K}}^{-1}$
  - $\mathrm{\Delta }\theta$ is the temperature change in $\mathrm{K}$
• The specific heat capacity will be given in the question. Do not use $c=4.20$ or $c=4.18$ from GCSE/A Level chemistry without converting the mass to $\mathrm{g}$ first, because these values are in joules per gram per kelvin

Specific latent heat:

• Melting a solid requires breaking the bonds between molecules. The amount of energy required to change the state of $1\mathrm{k}\mathrm{g}$ of the substance is called the specific latent heat of fusion/vaporisation, $L$
• $\mathrm{\Delta }E=L\mathrm{\Delta }m$ where
  - $\mathrm{\Delta }E$ is the energy change in $\mathrm{J}$
  - $L$ is the specific latent heat of fusion/vaporisation in $\mathrm{J}{\mathrm{k}\mathrm{g}}^{-1}$
  - $\mathrm{\Delta }m$ is the mass in $\mathrm{k}\mathrm{g}$

145 Using a thermistor and potential divider as a thermometer (CP 12):

• Set up a potential divider with a fixed resistor and waterproof NTC thermistor. Choose the fixed resistor so that ${\mathrm{V}}_{\mathrm{o}\mathrm{u}\mathrm{t}}$ will vary enough as temperature changes
  
• An NTC thermistor's resistance will decrease as temperature increases
• Use ice and a Bunsen burner to vary the temperature of a beaker of water. Record the temperature (with a thermometer) and output voltage
• Plot a calibration curve of temperature against voltage
• Now this curve can be used to find the temperature of a solution without a thermometer
• Alternatively, plot $\ln R$ against $\frac{1}{T}$, which is a straight line graph. Reading off a straight line graph is generally more accurate than from a curve, but it takes longer to plot

146 Determining the specific latent heat of a phase change (CP 13):

• We need to measure the energy change when ice melts (specific latent heat of fusion)
• Place ice in a funnel and allow it to warm up to 0 °C
• Add $100{\mathrm{c}\mathrm{m}}^3$ of water to a beaker and measure the mass and temperature
• Add ice at 0 °C to the beaker and stir until it all melts
• Record the lowest temperature the mixture reaches
• Weigh the beaker with melted ice and water
• Due to conservation of energy, we have the relationship
  energy cooling water $=$ energy melting ice $+$ energy warming melted ice
• With the two formulae in 144, you can derive this:
  $m_{\mathrm{w}\mathrm{a}\mathrm{t}\mathrm{e}\mathrm{r}}\times c\times ({\theta }_1-{\theta }_2)$ $=m_{\mathrm{i}\mathrm{c}\mathrm{e}}\times L+m_{\mathrm{i}\mathrm{c}\mathrm{e}}\times c\times ({\theta }_2$ [$-0$]$)$

147 Internal energy:

• Internal energy is the total amount of energy in a system
• internal energy $=$ total kinetic energy of the molecules $+$ total potential energy of the molecules
• Each molecule in a system will have a different amount of internal energy - it is randomly distributed

Ideal gases:

• An ideal gas is a model of how the molecules in a gas behave, where the molecules collide perfectly elastically and move completely randomly
• This is not fully accurate to reality, but the equations based on ideal gas theory are good approximations
• Ideal gas particles do not have potential energy, so internal energy $=$ total kinetic energy

Phase changes:

• When a substance is heated, the kinetic energy of its molecules increases
• When a substance is changing phase, its potential energy is changing but kinetic energy (and hence temperature) remains constant

148 Kelvins:

• To convert from Celcius to kelvins, add 273
• $0\mathrm{K}$ is the lowest possible temperature; absolute zero
• At absolute zero, all particles would have the minimum possible internal energy, so they would stop moving
• Average kinetic energy is proportional to absolute temperature - i.e. increasing energy increases temperature

149 Deriving the ideal gas pressure equation:

• Let's say we have a hollow cube with length $l$, containing $N$ molecules, each with mass $m$
• A molecule (call it Q) moves towards a wall (call it W) with velocity $u$
  - Therefore, it's momentum is $mu$
• The molecule collides elastically with W. After, it has a momentum of $-mu$
• Therefore, the change in momentum is $-2mu$
• Now calculate the time between collisions with the wall W; $t=$ $\frac{s}{u}$ $=$ $\frac{2l}{u}$
  - So there are $(\frac{2l}{u}{)}^{-1}$ $=$ $\frac{u}{2l}$ collisions per second
• So the rate of change of momentum is $-2mu\times$ $\frac{u}{2l}$
  - This simplifies to $-\frac{m{u}^2}{l}$
• force $=$ the rate of change of momentum (Newton's second law), so the force exerted by the wall W on the molecule Q is also $-\frac{m{u}^2}{l}$
• Due to Newton's third law, the force exerted on the wall will be the negative of this, $\frac{m{u}^2}{l}$
• Because all particles will have a different velocity, we can say that the total force of every molecule on the wall is $F=$ $\frac{m(u_0^2+u_1^2+...+u_n^2)}{l}$
• The mean square speed ($⟨u^2⟩$) is the mean of the squared speed of the particles (in the $x$ direction only). $⟨u^2⟩=$ $\frac{u_0^2+u_1^2+...+u_n^2}{N}$
• The two equations can be combined to give $F=$ $\frac{Nm⟨u^2⟩}{l}$
• pressure $=$ force $÷$ area $=$ $\frac{Nm⟨u^2⟩}{l}$ $÷l^2$
• This simplifies to $p=$ $\frac{Nm⟨u^2⟩}{V}$ (since $V=l^3$)

Applying this in all 3 directions:

• A gas molecule will move in three dimensions, we'll call the velocity components in each $u$, $v$ and $w$
• Using Pythagoras' theorem, we can calculate the overall speed, $c$: $c^2=u^2+v^2+w^2$
• This gives an overall mean square speed of $⟨c^2⟩=⟨u^2⟩+⟨v^2⟩+⟨w^2⟩$
• Molecules move randomly in all dimensions, so the mean square velocities are equal with lots of molecules. Therefore, we can simplify the above to $⟨c^2⟩=3⟨u^2⟩$
• If we rearrange this to $⟨u^2⟩=$ $\frac{⟨c^2⟩}{3}$ and substitute it into the equation above, we get
  $pV=\frac{1}{3}Nm⟨c^2⟩$ where
  - $p$ is the pressure in $\mathrm{N}{\mathrm{m}}^{-2}$
  - $V$ is the volume in ${\mathrm{m}}^3$
  - $N$ is the number of molecules
  - $m$ is the mass of each molecule in $\mathrm{k}\mathrm{g}$
  - $⟨c^2⟩$ is the mean square speed of the gas molecules in ${\mathrm{m}}^2{\mathrm{s}}^{-2}$
  (this formula is given in the formula booklet, but we also need to be able to derive it)

Assumptions made in the derivation:

• Particles have no volume
• There is no force between particles
• Particles don't collide with other particles
• The particle size is much smaller than the box size
• The impact time is zero
• Motion is completely random

150 Using $pV=NkT$:

• $pV=NkT$ where
  - $p$ is the pressure in $\mathrm{P}\mathrm{a}$ ($\mathrm{N}{\mathrm{m}}^{-2}$)
  - $V$ is the volume in ${\mathrm{m}}^3$
  - $N$ is the number of molecules in the gas
  - $k$ is the Boltzmann constant ($1.38\times {10}^{-23}$ $\mathrm{J}{\mathrm{K}}^{-1}$, in data sheet)
  - $T$ is the temperature of the gas in $\mathrm{K}$
  Note - this equation uses the number of molecules instead of the number of moles which is used in A Level Chemistry

151 Investigating the effect of pressure on volume (CP 14):

• Use a pump and Boyle's law apparatus, as shown below:
  
• Increase the pressure with the pump and record the volume of air by reading off the scale. Repeat this several times for different pressures
• Try to maintain a constant temperature by waiting about $30\mathrm{s}$ after taking a reading before pumping again
• The volume occupied by air should be reduced as pressure increases
• Plot $p$ against $\frac{1}{V}$. You should get a straight line

152 Deriving the kinetic energy per molecule equation:

• Ideal gases do not have potential energy, so internal energy $=$ total kinetic energy
• The equations $pV=\frac{1}{3}Nm⟨c^2⟩$ and $pV=NkT$ can be merged to get $\frac{1}{3}Nm⟨c^2⟩=NkT$
• The $N$s on each side cancel each other out, so we have $\frac{1}{3}m⟨c^2⟩=kT$
• Now multiply both sides by 1.5 to get $\frac{1}{2}m⟨c^2⟩=\frac{3}{2}kT$
• Because $E_{\mathrm{k}}=\frac{1}{2}m{v}^2$, $\frac{1}{2}m⟨c^2⟩$ is the average kinetic energy per molecule
• So average kinetic energy per molecule $=\frac{1}{2}m⟨c^2⟩=\frac{3}{2}kT$
• Therefore, in an ideal gas, kinetic energy is proportional to absolute temperature

153 Black body radiators:

• All objects above absolute zero will emit radiation; usually as infrared, and visible light at very high temperatures
• Black surfaces can both absorb and emit all wavelengths of electromagnetic radiation
• This is called black body radiation and the wavelength and intensity of emitted radiation varies depending only on the temperature of the black body
• Stars approximately behave as black bodies
• The radiation curve shows intensity against wavelength for different temperatures. Only a small region ($390$ to $700\mathrm{n}\mathrm{m}$) is visible light:
  
• So from this, we can see that hot stars emit mostly high-frequency and short-wavelength radiation (you don't need to memorise it, this bullet point is just demonstrating how to read a radiation curve for a black body radiator)

154 The Stefan-Boltzmann law equation:

• Luminosity, $L$, is the total energy emitted by a radiator per second
• It is related to a black body radiator's temperature with the Stefan-Boltzmann law
• This is $L=\sigma A{T}^4$ where
  - $L$ is the luminosity of the black body radiator in $\mathrm{W}$
  - $\sigma$ is the Stefan-Boltzmann constant, $5.67\times {10}^{-8}$ $\mathrm{W}{\mathrm{m}}^{-2}{\mathrm{K}}^{-4}$. This is given in the data sheet
  - $A$ is the surface area of the radiator in ${\mathrm{m}}^2$
  - $T$ is the surface temperature of the radiator in $\mathrm{K}$
• If you have a spherical black body radiator (e.g. a star), surface area can be calculated with $A=4\pi r^2$

155 Wien's law:

• The peak wavelength of a temperature, ${\lambda }_{\mathrm{m}\mathrm{a}\mathrm{x}}$, is the wavelength with the highest intensity of emitted radiation
• It is related to an object's temperature with Wien's law; ${\lambda }_{\mathrm{m}\mathrm{a}\mathrm{x}}T=2.898\times {10}^{-3}$ $\mathrm{m}\mathrm{K}$ where
  - $T$ is the surface temperature in $\mathrm{K}$
  - $2.898\times {10}^{-3}$ $\mathrm{m}\mathrm{K}$ is a constant

156 Luminosity and intensity:

• The intensity of a star, $I$, is the power per square metre that reaches Earth (far away stars appear darker), measured in $\mathrm{W}{\mathrm{m}}^{-2}$
• The distance-intensity relationship is an inverse square law with equation $I=$$\frac{L}{4\pi d^2}$ where $d$ is the distance from the star in $\mathrm{m}$

157 Astronomical distances:

• An astronomical unit ($\mathrm{A}\mathrm{U}$) is the mean distance between the Earth and Sun (Earth's orbit isn't completely circular so we use the mean distance)
• One light year ($\mathrm{l}\mathrm{y}$) is the distance that a photon would travel in one year in a vacuum

Parallax:

• When you move, distant objects will still appear to be in the same position, whilst close ones appear to move. This is the parallax effect
• Because the position of Earth is changing due to its orbit, the angle to a star can be measured two times on different days (by comparing it to the positions of further away stars in the background) and parallax can be used to calculate its distance
• The diagram below shows how the angle of parallax is measured
  
• Because the Earth-Sun distance forms the adjacent and star-Sun distance forms the opposite side of a right-angled triangle, $\tan$ may be useful

158 Standard candles:

• Trigonometric parallax is only accurate for measuring the distance of closer stars
• For those further away, standard candles can be used. These are objects (e.g. supernovae) with a known luminosity
• By measuring the intensity of a standard candle on Earth, the inverse square law (see 156) can be used to calculate the distance to the standard candle. This allows astronomers to estimate the distance to any systems near to the candle

159 Hertzsprung-Russell (H-R) diagrams:

• Star luminosity can be plotted against surface temperature
• There is a correlation between the luminosity and surface temperature of stars
• Different types of star appear in different areas
• There are very few outliers because the transition period between different stages of a star's life is very quick
• If you are shown a H-R diagram of a star cluster, you can assume that they all formed at the same time
• A H-R diagram for nearby stars is shown below. You need to memorise everything on it (it is common for exam questions to ask you to fill in sections). All information is also listed in text form if you prefer that
  
  - The scale on the $y$ axis is a log scale of luminosity, relative to the sun, from ${10}^{-4}$ to ${10}^6$
  - The scale on the $x$ axis is the temperature in $\mathrm{K}$, $40000$ on the left, $10000$ in the centre and $2500$ on the right
  - The Sun has a surface temperature of $6000\mathrm{K}$
  - White dwarfs are in the bottom left corner
  - Blue giants are in the top left corner
  - Red supergiants are in the top right corner, with red giants slightly below and left of them
  - The main sequence stars form a straight line from $(40000,{10}^4)$ to $(2500,{10}^{-4})$

160 New stars:

• Clouds of dust and gas from supernovae contract due to gravity. The density continues to increase and it heats up. At this stage, it is slowly moving from the right of the H-R diagram towards the main sequence
• Once fusion of hydrogen to helium starts in the core, the star is in the main sequence
• Large stars are bright and hot, at the top left of the main sequence
• Small stars are dim and cooler, at the bottom right of the main sequence
• Larger stars will use up their fuel more quickly and spend less time in the main sequence. Therefore, stars at the top left of the main sequence will become red giants first
• So a very old cluster will have no stars at the top left, and lots of red giants and white dwarfs (both of which are formed after a star leaves the hydrogen fusion stage)

Red giants:

• Once the hydrogen in a star has been used up, nuclear fusion can no longer continue
• The core contracts and heats up. Helium starts to fuse and the outer layers cool and expand
• This stage is a red giant (or red supergiant)

White dwarfs:

• Once all of the helium has been used up from fusion in the red giant stage, the star again starts to collapse and no more fusion can occur. It will eventually reach a stable size
• As the outer layers become more unstable, they are ejected into space as nebulae
• The remaining core is hot and dense. This is a white dwarf

Supernovae, black holes and neutron stars:

• Large red giants keep fusing after all helium is used up
• Finally, the star explodes in a supernova
• The core is left behind, forming a neutron star or black hole depending on its mass
• Neutron stars are so small that their luminosity is off the scale of a H-R diagram
• Black holes don't emit any light, so they also aren't seen on H-R diagrams

161 The Doppler effect:

• If a moving object is emitting waves in all directions, the wavelength will be shorter in front of it and longer behind it
• This is because each crest is emitted closer to the observer than the previous (since the emitter is moving). Since more cycles are received per second by the observer, the perceived frequency will be higher
• Due to $v=f\lambda$, this means that the frequency is higher in front and lower behind
• The change in frequency is called the Doppler shift, and higher speeds result in a greater 'shift'

162 Redshift:

• If a light source (e.g. a star) is moving away, the wavelength we observe is longer due to the Doppler effect (see above) - this causes the light to shift further towards the red end of the visible spectrum. This is called redshift
• The opposite occurs if a source is getting closer to the observer; this is called blueshift
• Provided that $v$ is much less than $c$, $z=$ $\frac{\mathrm{\Delta }\lambda }{\lambda }$ $\approx$ $\frac{\mathrm{\Delta }f}{f}$ $\approx$ $\frac{v}{c}$ where
  - $z$ is the amount of redshift/blueshift
     - if $z>0$, the observer sees redshift
     - if $z<0$, the observer sees blueshift
  - $\lambda$ is the emitted wavelength, in $\mathrm{m}$
  - $\mathrm{\Delta }\lambda$ is the difference between the observed and emitted wavelength (${\lambda }_{\mathrm{o}\mathrm{b}\mathrm{s}\mathrm{e}\mathrm{r}\mathrm{v}\mathrm{e}\mathrm{d}}-{\lambda }_{\mathrm{e}\mathrm{m}\mathrm{i}\mathrm{t}\mathrm{t}\mathrm{e}\mathrm{d}}$)
  - $f$ is the emitted wavelength, in $\mathrm{H}\mathrm{z}$
  - $\mathrm{\Delta }f$ is the difference between the observed and emitted wavelength ($f_{\mathrm{e}\mathrm{m}\mathrm{i}\mathrm{t}\mathrm{t}\mathrm{e}\mathrm{d}}-f_{\mathrm{o}\mathrm{b}\mathrm{s}\mathrm{e}\mathrm{r}\mathrm{v}\mathrm{e}\mathrm{d}}$)
  - $v$ is the velocity of the object relative to the observer
     - If it's moving further away from the observer, $v$ is positive
     - If it's moving closer to the observer, $v$ is negative
  - $c$ is the speed of light

Hubble's law:

• Recessional velocity is the rate at which an astronomical object is moving away from Earth
• The recessional velocity of galaxies is proportional to how far they are away. They are related in the equation $v=H_{0}d$, where
  - $v$ is the recessional velocity in $\mathrm{k}\mathrm{m}{\mathrm{s}}^{-1}$
  - $H_0$ is Hubble's constant (about $72\mathrm{k}\mathrm{m}{\mathrm{s}}^{-1}{\mathrm{M}\mathrm{p}\mathrm{c}}^{-1}$)
  - $d$ is the distance in $\mathrm{M}\mathrm{p}\mathrm{c}$

163 Controversy over the Hubble constant:

• Assuming the rate of expansion of the universe is constant (although we have some evidence that it isn't), its age can be calculated with $t=$ $\frac{\mathrm{d}\mathrm{i}\mathrm{s}\mathrm{t}\mathrm{a}\mathrm{n}\mathrm{c}\mathrm{e}}{\mathrm{s}\mathrm{p}\mathrm{e}\mathrm{e}\mathrm{d}}$ $=$ $\frac{1}{H_0}$ (if $H_0$ is converted to ${\mathrm{s}}^{-1}$)
• However, we only know the value of $H_0$ is about 72; different calculation methods produce a different value and hence give a different approximation of the age of the universe
• Also, the size of the universe is unknown; we can only see matter that has been able to travel to Earth at the speed of light

The fate of the universe and dark matter:

• Depending on its density, the universe may or may not keep expanding forever
• However, density is difficult to calculate due to evidence of dark matter
• Dark matter is matter that isn't visible but has a mass

164 Nuclear binding energy:

• If you sum the expected masses of the protons and neutrons in an atom, it will be more than the actual mass of the nucleus
• This mass difference is called the mass deficit
• When protons and neutrons join, the mass decreases and is released as energy
• The mass deficit can be calculated with $\mathrm{\Delta }E=c^2\mathrm{\Delta }m$, where
  - $m$ is the deficit
  - $\mathrm{\Delta }E$ is the energy released
• The nuclear binding energy is the energy needed to separate all of the protons and neutrons in a nucleus. This is equal to the mass deficit and is the amount of energy released in a fusion reaction
• Binding energy is usually measured in $\mathrm{M}\mathrm{e}\mathrm{V}$

165 Atomic mass units:

• The atomic mass unit, $\mathrm{u}$, is a unit of mass. It is equal to ${\frac{1}{12}}^{\mathrm{t}\mathrm{h}}$ of the mass of a ${}^{12}\mathrm{C}$ atom
• Since electrons have a very low mass compared to nuclei, $\mathrm{u}$ is very close to the mass of a proton/neutron
• To convert from $\mathrm{k}\mathrm{g}$ to $\mathrm{u}$, divide the mass (in $\mathrm{k}\mathrm{g}$) by $1.66\times {10}^{-27}$ (given in the Data Sheet)
• To convert from $\mathrm{u}$ to $\mathrm{k}\mathrm{g}$, multiply the mass (in $\mathrm{u}$) by $1.66\times {10}^{-27}$

166 Nuclear fission and fusion:

• Nuclear fission is where nuclei are split into smaller parts, releasing energy and sometimes small particles like neutrons (${}_0^{1}n$)
• Nuclear fusion is where two or more small nuclei fuse together to form a larger atomic nucleus and sometimes other subatomic particles like neutrons

Binding energy per nucleon:

• This can be used to compare binding energies. It usually has the units $\mathrm{M}\mathrm{e}\mathrm{V}$
• Calculate binding energy per nucleon by dividing the binding energy by the nucleon number (the number of protons and neutrons)
• Plotting binding energy per nucleon against nucleon number produces the below curve:
  
• The part on the left with the increasing gradient is the atoms where fusion occurs. After fusion, the binding energy is higher, so energy is released
• Fission occurs for the elements where the gradient is decreasing. As with fusion, the total binding energy after the process is higher than before, so energy is released
• You can calculate the energy released in fission/fusion by subtracting the total binding energy to the left of the → from the total binding energy to the right of the →

167 Requirements for nuclear fusion:

• Nuclei have a positive charge, so there will be an electrostatic force of repulsion between them
• For fusion to occur, the nuclei need to overcome this force, which requires a large amount of energy
• Therefore, fusion requires high temperatures
• Additionally, high pressure will increase the frequency of collisions and therefore rate of fusion
• The energy released helps maintain the high temperature for the fusion to continue with other nearby nuclei

168 Background radiation:

• Rocks contain radioactive isotopes and therefore emit radiation. Radon gas from rocks is also radioactive
• Cosmic rays (protons) colliding with atmospheric gases produce radiation
• Living things contain some ${}^{14}\mathrm{C}$ which is radioactive
• Industry also emits some radiation
• Therefore, whenever you are making measurements of radioactive sources, you must first measure the background radiation. Subtract this from all subsequent measurements
• It is best to record the number of counts of background radiation for at least a few minutes, then divide the count by the number of seconds you were measuring for

169 Nuclear radiation:

• Ionising radiation is radiation which can 'knock off' electrons from atoms, creating an ion (a charged atom)
• Radiation that has a charge can cause ionisation
• Radiation loses energy (and speed) every time it ionises an atom. Because alpha particles are heavier than beta (electrons), they have more energy, allowing them to ionise more atoms. They are also much larger, so they ionise a higher proportion of the atoms they pass by. Because they are losing energy so fast, they aren't very penetrating
• There are 4 types of nuclear radiation. Their properties are listed below (this is probably just a GCSE recap, with the exception of 'beta-plus', which is the antimatter equivalent of 'beta'/'beta-minus')

The 4 types of nuclear radiation:

• Alpha radiation ($\alpha$):
  - helium nuclei (2 protons, 2 neutrons)
  - $+2$ relative charge
  - $4\mathrm{u}$ mass
  - high ionising ability
  - low speed (about ${10}^7\mathrm{m}{\mathrm{s}}^{-1}$)
  - absorbed by a few sheets of paper, or a few $\mathrm{c}\mathrm{m}$ of air
  - deflected by magnetic fields
• Beta-minus radiation (${\beta }^{-}$):
  - electrons
  - low ionising ability
  - very fast (close to the speed of light)
  - absorbed by about $3\mathrm{m}\mathrm{m}$ of aluminium
  - deflected by magnetic fields
• Beta-plus radiation (${\beta }^{+}$):
  - positrons (anti-electrons)
  - because they're annihilated by electrons, they don't last very long
• Gamma radiation ($\gamma$):
  - short-wavelength, high-frequency electromagnetic waves/photons
  - very little ionising ability
  - speed of light
  - chargeless and massless
  - absorbed by a few $\mathrm{c}\mathrm{m}$ of lead, a few $\mathrm{m}$ of concrete
  - not deflected by magnetic fields

170 Nuclear decay equations:

• When writing particles, use the notation ${}_Z^{A}X$, where
  - $A$ is the nucleon number ($A$tomic mass of the nucleus)
  - $Z$ is the proton number (number of proton$Z$)
  - $X$ is the symbol that corresponds to the proton number on the periodic table (e.g. $\mathrm{A}\mathrm{m}$ if you have $95$ protons)
• In other words, mass (usually the largest of the two numbers) is on top, number of protons is on the bottom
• The proton number is always equal to the charge in nuclear equations. For example, beta-minus is written as ${}_{-1}^0\beta$
• Equations must be balanced in terms of:
  - charge (the sum of proton numbers must be equal on both sides of the →)
  - nucleon number (the sum of nucleon numbers must be equal on both sides of the →)
• Energy and momentum are always conserved
• Mass doesn't have to be conserved - some can be released as energy

Alpha decay:

• Alpha particles are only emitted from very heavy atoms (e.g. $\mathrm{U}$)
• They can be written as ${}_2^4\alpha$

Beta-minus decay:

• Beta decay happens to isotopes which are neutron-rich (mostly made up of neutrons)
• One neutron from the nucleus decays, turning into a proton. This releases:
  - a beta-minus particle (electron) (${}_{-1}^0\beta$)
  - an antineutrino (${}_0^0\overline{\nu }$)

Beta-plus decay:

• Beta-plus decay is where a proton changes to a neutron (the opposite of beta-minus), releasing:
  - a beta-plus particle (positron) (${}_1^0\overline{\beta }$)
  - a neutrino (${}_0^0\nu$)

Gamma emission:

• Very excited isotopes (with lots of excess energy) increase their stability by losing energy in the form of a gamma ray photon
• After alpha/beta decay, gamma emission sometimes also occurs
• In nuclear equations, gamma can be written as $\gamma$ or ${}_0^0\gamma$

171, 172 Investigating the absorption of gamma radiation by lead (CP15):

• Measure the background count rate
• Now place a clamp between a gamma source and a Geiger counter, with about $15\mathrm{c}\mathrm{m}$ between them
• Record the number of counts in one minute. Divide by 60 to get the count rate
• Place different thicknesses of lead into the clamp. For each, measure the thickness with a micrometer. Record the count rate with each thickness and subtract the background count rate to get the corrected count rate
• Plot $\ln ($corrected count rate$)$ against lead thickness
• The half-value thickness (thickness of lead required to block half of the gamma radiation) is $\ln 2÷$ gradient
• Since radiation is random, there will be some uncertainty in the results

173 Radioactive decay equations:

• The decay constant, $\lambda$, is the probability of a nucleus decaying per second, measured in ${\mathrm{s}}^{-1}$
• $N$ is the number of undecayed nuclei remaining
• The activity, $A$, is the number of nuclei that decay per second. It is measured in becquerels ($\mathrm{B}\mathrm{q}$), where $1\mathrm{B}\mathrm{q}=1\mathrm{n}\mathrm{u}\mathrm{c}\mathrm{l}\mathrm{e}\mathrm{u}\mathrm{s}\mathrm{d}\mathrm{e}\mathrm{c}\mathrm{a}\mathrm{y}\mathrm{p}\mathrm{e}\mathrm{r}\mathrm{s}\mathrm{e}\mathrm{c}\mathrm{o}\mathrm{n}\mathrm{d}$
• $t_{\frac{1}{2}}$ is the half-life; the time it takes for half of the nuclei to decay. It is measured in seconds/minutes/hours/days etc
• $A=\lambda N$
• $-\lambda N=$ $\frac{\mathrm{d}N}{\mathrm{d}t}$
• $A=A_0{\mathrm{e}}^{-\lambda t}$ where $A_0$ is the initial activity and $A$ is the activity after $t$ seconds
• $N=N_0{\mathrm{e}}^{-\lambda t}$ where $N_0$ is the initial number of undecayed nuclei and $N$ is the number of undecayed nuclei after $t$ seconds
• Half-life and the decay constant are related by the equation $\lambda =$$\frac{\ln 2}{t_{1/2}}$

174 Gravitational fields:

• A gravitational field is a region in which objects with mass will experience a force

175 Gravitational field strength:

• $g=$$\frac{F}{m}$ where
  - $g$ is the gravitational field strength ($\mathrm{N}{\mathrm{k}\mathrm{g}}^{-1}$)
  - $F$ is the force on the mass ($\mathrm{N}$)
  - $m$ is the mass ($\mathrm{k}\mathrm{g}$)
• This can be rearranged to $F=mg$. This is similar to $F=ma$, which is why $g$ is also the acceleration due to gravity of a mass in a gravitational field
• $g$ is a vector, pointing towards the centre of mass of the object causing the gravitational field

176 Newton's law of universal gravitation:

• A small mass in a gravitational field is attracted to the mass causing the field and vice-versa - due to Newton's third law, the force each way is equal
• A point mass is an object with a mass but no dimensions (i.e. it's a 'point'). It cannot actually occur, but the concept can be used to simplify a spherical mass (e.g. a planet)
• Newton's law of universal gravitation is $F=$ $\frac{G{m}_1{m}_2}{r^2}$ where
  - $F$ is the force acting on mass 1 due to mass 2
  - $G$ is the gravitational constant ($6.67\times {10}^{-11}$ $\mathrm{N}{\mathrm{m}}^2{\mathrm{k}\mathrm{g}}^{-2}$, given in the data sheet)
  - $m_1$ is the mass of mass 1 in $\mathrm{k}\mathrm{g}$
  - $m_2$ is the mass of mass 2 in $\mathrm{k}\mathrm{g}$
  - $r$ is the distance between the centre of mass of mass 1 and the centre of mass of mass 2 in $\mathrm{m}$
     - For near-spheres like Earth, we can assume that the centre of mass is in the centre of the planet

177 $g$ and $r$ are related by an inverse square relationship:

• In the gravitational field of $m_2$, $g=$ $\frac{F}{m_1}$ (see 175)
• Also, $F=$ $\frac{G{m}_1{m}_2}{r^2}$ (see 176)
• Rearrange the first equation to $F=g{m}_1$
• Now the two equations can be equated; $g{m}_1=$ $\frac{G{m}_1{m}_2}{r^2}$
• Now divide both sides by $m_1$: $g=$ $\frac{G{m}_2}{r^2}$
• $m_2$ is causing the gravitational field, so this can be written as $g=$ $\frac{Gm}{r^2}$ where
  - $m$ is the mass of the object creating the field ($\mathrm{k}\mathrm{g}$)
  - $g$ is the gravitational field strength ($\mathrm{N}{\mathrm{k}\mathrm{g}}^{-1}$)
  - $G$ is the gravitational constant (see 176)
  - $r$ is the distance between the two centres of mass
• This equation shows that the gravitational field strength is inversely proportional to $r^2$ - i.e. an inverse square law
• The equation works for the area around a uniform sphere, but not inside one. Inside, $g\propto r$ (a linear relationship)

178 Gravitational potential:

• The gravitational potential at a point is the gravitational potential energy per unit mass; i.e. the work that would need to be done to bring a unit mass from infinity to that point
• It is always negative because gravitational fields are attractive to all objects with a mass
• It is a scalar quantity (since energy is a scalar quantity)
• So on Earth, gravitational potential is most negative at the surface, increasing as altitude increases, converging to 0
• If this is confusing you due to the mechanics we were taught at GCSE/AS, remember that we used (or should've used) $\mathrm{\Delta }\mathrm{G}\mathrm{P}\mathrm{E}$, not just '$\mathrm{G}\mathrm{P}\mathrm{E}$'
• $V_{\mathrm{g}\mathrm{r}\mathrm{a}\mathrm{v}}=$ $\frac{-GM}{r}$ where
  - $V_{\mathrm{g}\mathrm{r}\mathrm{a}\mathrm{v}}$ is the gravitational potential ($\mathrm{J}{\mathrm{k}\mathrm{g}}^{-1}$)
  - $G$ is the gravitational constant
  - $M$ is the mass of the object creating the field ($\mathrm{k}\mathrm{g}$)
  - $r$ is the distance from the centre of the object creating the field ($\mathrm{m}$)

179 Similarities between electric fields and gravitational fields:

• Potential energy:
  - Gravitational potential ($V_{\mathrm{g}\mathrm{r}\mathrm{a}\mathrm{v}}$) is the potential energy per unit mass
  - Electric potential ($V$) is the potential energy per unit positive charge
• Field strength:
  - Gravitational field strength ($g$) is the force per unit mass
  - Electric field strength ($E$) is the force per unit positive charge
• Point masses/charges:
  - The force between two point masses with different distances is an inverse square law (Newton's law of universal gravitation)
  - The force between two point charges with different distances is an inverse square law (Coulomb's law)
• Field lines:
  - Gravitational field lines point towards the object creating the field and are closer together closer to the object
  - Electric field lines for a negative point charge point towards the charge and are closer together closer to the point charge

Differences between gravitational fields and electric fields:

• Attraction:
  - Gravitational forces are always attractive
  - Electric forces can be repulsive or attractive
• The medium:
  - Electric forces depend on the medium between the charges, gravitational forces do not (the medium is what's between them - i.e. vacuum, air)
• Shielding:
  - Objects can only be shielded from electric fields

180 Orbital motion:

• An orbiting body has a constant speed. But since it's constantly changing direction, it's accelerating
• All accelerating objects experience a force (Newton's first law), this force is centripetal force
• The centripetal force is $F=$ $\frac{m_1{v}^2}{r}$
• This centripetal force must be caused by something (conservation of energy); this is gravitational force
• Therefore, centripetal force must be equal to gravitational force: $\frac{m_1{v}^2}{r}$ $=$ $\frac{G{m}_1{m}_2}{r^2}$
• This simplifies to $v=$ $\sqrt{\frac{GM}{r}}$ where
  - $v$ is the orbital speed ($\mathrm{m}{\mathrm{s}}^{-1}$)
  - $G$ is the gravitational constant
  - $M$ is the mass of the body that is creating the gravitational field ($\mathrm{k}\mathrm{g}$)
  - $r$ is the distance between the centres of mass of the two bodies ($\mathrm{m}$)
  Note: you'll have to derive this in the exam, it's not in the formula booklet
• The period of the orbit, $T$ can be calculated with $T=$ $\frac{2\pi }{\omega }$ and $v=\omega r$

181 Simple harmonic motion:

• In simple harmonic motion, there is a restoring force ($F$) pushing the object towards the midpoint
• Farther from the midpoint, the restoring force is greater
• Displacement (from the midpoint) and restoring force are related in the formula $F=-kx$ where
  - $F$ is the restoring force in $\mathrm{N}$
  - $k$ is the spring constant for the spring
  - $x$ is the displacement from the midpoint in $\mathrm{m}$
• Any oscillation which fits the above formula is considered SHM
• The formula contains a $-$ sign because the restoring force is always acting in the direction opposite to the displacement

182 Equations:

• The period, $T$, is the time taken for one oscillation (in $\mathrm{s}$)
• The frequency, $f$, is the number of oscillations per second in $\mathrm{H}\mathrm{z}$ or ${\mathrm{s}}^{-1}$
• Frequency and period are independent of the amplitude
• The angular frequency, $\omega$, is the number of radians of the oscillation per second (one oscillation is one full circle, $2\pi \mathrm{r}\mathrm{a}\mathrm{d}$). It is measured in $\mathrm{r}\mathrm{a}\mathrm{d}{\mathrm{s}}^{-1}$ and calculated with the formula $\omega =2\pi f$
• $x=A\cos (\omega t)$ where $A$ is the amplitude of the oscillation (the maximum displacement from the midpoint)
• $v=-A\omega \sin (\omega t)$
• $a=-{\omega }^{2}x=-A{\omega }^2\cos (\omega t)$
• $T=$ $\frac{2\pi }{\omega }$ $=$ $\frac{1}{f}$

183 More equations:

• For a mass-spring system (which is SHM), the period, mass and spring constant are related by $T=2\pi$$\sqrt{\frac{m}{k}}$ where
  - $T$ is the period in $\mathrm{s}$
  - $m$ is the mass of the oscillating object in $\mathrm{k}\mathrm{g}$
  - $k$ is the spring constant of the spring in $\mathrm{N}{\mathrm{m}}^{-1}$
• A pendulum is also a simple harmonic oscillator, the formula $T=2\pi$$\sqrt{\frac{l}{g}}$ applies where
  - $T$ is the period in $\mathrm{s}$
  - $l$ is the length of the pendulum in $\mathrm{m}$
  - $g$ is the gravitational field strength in $\mathrm{N}{\mathrm{k}\mathrm{g}}^{-1}$

184 Displacement-time graphs:

• If displacement is plotted against time, you get a graph of $x=A\cos (\omega t)$ [this is slightly confusing because displacement ($x$), is plotted on the $y$ axis]
• The gradient at a point gives the velocity

185 Velocity-time graphs:

• Velocity is $\frac{1}{4}$ of a cycle in front of the displacement (i.e. translated $\frac{\pi }{2}$ to the left)
• The equation of the graph is $v=-A\omega \sin (\omega t)$
• The gradient at a point gives the acceleration

186 Natural frequency:

• The natural frequency of an oscillator is the frequency it will oscillate at (with no driving force)
• It is independent of amplitude; every object will have a natural frequency
• For example, if you extend a spring and then release it, and it starts oscillating at $3\mathrm{H}\mathrm{z}$, then its natural frequency is $3\mathrm{H}\mathrm{z}$

Resonance:

• If a force is being applied to an oscillator, the amplitude increases as the frequency of the driving force approaches the natural frequency of the oscillator as more energy is being transferred
• At this peak energy transfer, the system is said to be resonating at its resonating frequency

187 Using SHM to calculate the mass of an object (CP 16):

• Record the period for 10 oscillations with a spring and varying masses (e.g. $100\mathrm{g}$, $200\mathrm{g}$, ..., $500\mathrm{g}$). Do three repeats for each mass
• Many school masses will have an uncertainty of about 5%, so it's best to weigh each and record that mass
• Plot a straight line graph of period ($T$) against the square root of the mass ($\sqrt{m}$)
• Now record the period for the unknown mass. Use the calibration graph you created to determine its mass

188 Damping:

• If an oscillator isn't damped, no energy will be transferred to the surroundings, so it will keep oscillating forever due to conservation of energy
• With damping, energy will be lost to the surroundings (e.g. as heat). But energy is still conserved

189 Free oscillations:

• If there is no external force acting on an oscillator and it is not losing any energy to the surroundings, it is called a free oscillation
• This is impossible in reality, but sometimes the energy loss is negligible, such as with some springs oscillating in air

Forced oscillations:

• A system can be forced to oscillate by an external force
• The frequency of this force is the driving frequency

190 The effect of resonance on amplitude:

• As the driving frequency approaches the natural frequency, the amplitude increases quickly. This is shown below:
  
• The peak is flatter in heavily damped systems
• The resonant frequency of a damped oscillator is lower than its natural frequency because it is being slowed by the damper

191 The effect of damping on amplitude:

• The amplitude of a damped oscillator will decrease over time
• The rate of this will reduce with time (so it would take longer for it the amplitude to go from $2\mathrm{m}$ to $1\mathrm{m}$ than it would to reduce from $4\mathrm{m}$ to $3\mathrm{m}$

Ductile materials:

• A ductile material is defined as a material that can be easily stretched into a thin wire. As a damper, a more ductile material will be able to undergo plastic deformation more easily
• Plastic deformation is irreversible, so it is an effective way to absorb energy