Key Equations

Turning Points in Physics | AQA A-Level Physics

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12.1: The Discovery of the Electron
Work done on an electron
$$W = eV$$
  • Where:
    • $W$ = work done / kinetic energy gained (J)
    • $e$ = charge of the electron (C)
    • $V$ = accelerating potential difference (V)
Equates to $\frac{1}{2}m_e v^2$ when the electron starts from rest.
Specific charge (magnetic field only)
$$\frac{e}{m_e} = \frac{2V_a}{B^2 r^2}$$
  • Where:
    • $V_a$ = accelerating p.d. (V)
    • $B$ = magnetic flux density (T)
    • $r$ = radius of circular path (m)
Derived by equating $eV_a = \frac{1}{2}m_e v^2$ and $Bev = m_e v^2 / r$.
Millikan's stationary drop condition
$$Q = \frac{mgd}{V}$$
  • Where:
    • $Q$ = charge on the oil drop (C)
    • $m$ = mass of the drop (kg)
    • $g$ = gravitational field strength (N kg$^{-1}$)
    • $d$ = plate separation (m)
    • $V$ = p.d. across plates (V)
Valid only when the drop is stationary ($QE = mg$).
Thomson's crossed-fields method
$$\frac{e}{m_e} = \frac{V}{B^2 r d}$$
  • Where:
    • $V$ = p.d. across deflecting plates (V)
    • $B$ = magnetic flux density (T)
    • $r$ = radius of curvature in B field alone (m)
    • $d$ = plate separation (m)
Uses $v = V/(Bd)$ from balanced fields, then $e/m_e = v/(Br)$.
Stokes' law
$$F = 6\pi \eta r v$$
  • Where:
    • $F$ = viscous drag force (N)
    • $\eta$ = viscosity of the fluid (Pa s)
    • $r$ = radius of the sphere (m)
    • $v$ = speed of the sphere (m s$^{-1}$)
At terminal velocity, $mg = 6\pi \eta r v_t$. Used to find the oil drop radius.
Oil drop radius from terminal velocity
$$r = \sqrt{\frac{9 \eta v_t}{2 \rho g}}$$
  • Where:
    • $\eta$ = viscosity of air (Pa s)
    • $v_t$ = terminal velocity of the drop (m s$^{-1}$)
    • $\rho$ = density of oil (kg m$^{-3}$)
    • $g$ = gravitational field strength (N kg$^{-1}$)
Derived from $\frac{4}{3}\pi r^3 \rho g = 6\pi \eta r v_t$. Mass then follows from $m = \frac{4}{3}\pi r^3 \rho$.
12.2: Wave-Particle Duality
De Broglie wavelength
$$\lambda = \frac{h}{p} = \frac{h}{mv}$$
  • Where:
    • $\lambda$ = de Broglie wavelength (m)
    • $h$ = Planck constant ($6.63 \times 10^{-34}$ J s)
    • $p$ = momentum (kg m s$^{-1}$)
All matter has an associated wavelength. Wavelength decreases with increasing momentum.
Fizeau's speed of light
$$c = 4dnf$$
  • Where:
    • $d$ = distance from wheel to mirror (m)
    • $n$ = number of teeth on the wheel
    • $f$ = rotation frequency at first eclipse (Hz)
At first eclipse, the wheel rotates by half a tooth gap while the light completes the round trip $2d$.
Speed of light from Maxwell's equation
$$c = \frac{1}{\sqrt{\mu_0 \varepsilon_0}}$$
  • Where:
    • $c$ = speed of light in a vacuum (m s$^{-1}$)
    • $\mu_0$ = permeability of free space (H m$^{-1}$)
    • $\varepsilon_0$ = permittivity of free space (F m$^{-1}$)
Maxwell showed that the speed of electromagnetic waves equals the measured speed of light.
Planck's quantisation
$$E = nhf$$
  • Where:
    • $E$ = energy of the oscillator (J)
    • $n$ = integer (quantum number)
    • $h$ = Planck constant (J s)
    • $f$ = frequency of oscillation (Hz)
Resolved the ultraviolet catastrophe. Energy is emitted in discrete packets (quanta).
De Broglie wavelength of an accelerated electron
$$\lambda = \frac{h}{\sqrt{2m_e eV}}$$
  • Where:
    • $V$ = accelerating potential difference (V)
    • $m_e$ = electron mass (kg)
    • $e$ = electron charge (C)
Combines $eV = \frac{1}{2}m_e v^2$ with $\lambda = h/(m_e v)$. Used for electron diffraction calculations.
12.3: Special Relativity
Lorentz factor
$$\gamma = \frac{1}{\sqrt{1 - v^2/c^2}}$$
  • Where:
    • $\gamma$ = Lorentz factor (dimensionless, always $\geq 1$)
    • $v$ = speed of the object (m s$^{-1}$)
    • $c$ = speed of light ($3.00 \times 10^8$ m s$^{-1}$)
At everyday speeds, $\gamma \approx 1$. At $v = 0.98c$, $\gamma \approx 5$.
Relativistic kinetic energy
$$E_k = (\gamma - 1)\,m_0 c^2$$
  • Where:
    • $E_k$ = relativistic kinetic energy (J)
    • $\gamma$ = Lorentz factor
    • $m_0$ = rest mass (kg)
    • $c$ = speed of light (m s$^{-1}$)
At low speeds, $\gamma - 1 \approx v^2/(2c^2)$, giving $E_k \approx \frac{1}{2}m_0 v^2$.
Time dilation
$$\Delta t = \gamma \Delta t_0$$
  • Where:
    • $\Delta t$ = dilated time measured by external observer (s)
    • $\Delta t_0$ = proper time in the moving frame (s)
    • $\gamma$ = Lorentz factor
Moving clocks run slower. $\Delta t_0$ is always the shorter time interval.
Relativistic mass
$$m = \gamma m_0$$
  • Where:
    • $m$ = relativistic mass (kg)
    • $\gamma$ = Lorentz factor
    • $m_0$ = rest mass (kg)
As $v \to c$, $\gamma \to \infty$, so $m \to \infty$. This is why $c$ can never be reached.
Length contraction
$$L = \frac{L_0}{\gamma}$$
  • Where:
    • $L$ = contracted length measured by external observer (m)
    • $L_0$ = proper length in the rest frame (m)
    • $\gamma$ = Lorentz factor
Objects moving relative to an observer are shorter in the direction of motion. $L$ is always $\leq L_0$.
Mass-energy equivalence
$$E = mc^2$$
  • Where:
    • $E$ = total energy (J)
    • $m$ = relativistic mass $= \gamma m_0$ (kg)
    • $c$ = speed of light (m s$^{-1}$)
Total energy = rest energy + kinetic energy: $E = m_0 c^2 + E_k$.
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