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Paper 2 Answers

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SetSet 4

PaperLevel 1 - Physics Paper 2

Information

Section A: Electric Fields
Section B: Electromagnetism
Section C: Relativity
Section D: Quantum Physics

Marking Guidance

These worked answers show the expected method and final result. Equivalent correct reasoning should receive credit.

Constants

Gravitational acceleration	\( g=9.81\,\mathrm{m\,s^{-2}} \)
Speed of light	\( c=3.00\times10^8\,\mathrm{m\,s^{-1}} \)
Elementary charge	\( e=1.60\times10^{-19}\,\mathrm{C} \)
Electron mass	\( m_e=9.11\times10^{-31}\,\mathrm{kg} \)
Planck constant	\( h=6.63\times10^{-34}\,\mathrm{J\,s} \)
Permittivity of free space	\( \epsilon_0=8.85\times10^{-12}\,\mathrm{F\,m^{-1}} \)
Magnetic constant	\( \mu_0=1.26\times10^{-6}\,\mathrm{N\,A^{-2}} \)
Boltzmann constant	\( k_B=1.38\times10^{-23}\,\mathrm{J\,K^{-1}} \)

Section A: Electric Fields

Answer 1. Three-charge field mapping on a line

[15 marks]

Three point charges are fixed on the \(x\)-axis. A charge \(+3.00\,\mathrm{nC}\) is at \(x=-0.300\,\mathrm{m}\), a charge \(-6.00\,\mathrm{nC}\) is at the origin, and a charge \(+3.00\,\mathrm{nC}\) is at \(x=+0.300\,\mathrm{m}\). Point \(P\) is at \(x=+0.150\,\mathrm{m}\). Take \(V(\infty)=0\).

a) Calculate the electric potential at \(P\).

[3 marks]

Electric potential adds as a scalar: \[ V_P=k\sum_i\frac{q_i}{r_i}. \] At \(P\), the distances from the left, middle, and right charges are \(0.450\,\mathrm{m}\), \(0.150\,\mathrm{m}\), and \(0.150\,\mathrm{m}\). Therefore \[ V_P=(8.99\times10^9)\left(\frac{3.00\times10^{-9}}{0.450}-\frac{6.00\times10^{-9}}{0.150}+\frac{3.00\times10^{-9}}{0.150}\right) \] \[ V_P=-1.20\times10^2\,\mathrm{V}. \]

Answer 1. Three-charge field mapping on a line

[15 marks]

b) Calculate the electric field at \(P\), giving its direction along the axis.

[4 marks]

Take \(+x\) as positive. The left positive charge produces a field in the \(+x\) direction at \(P\): \[ E_L=\frac{(8.99\times10^9)(3.00\times10^{-9})}{0.450^2}=1.33\times10^2\,\mathrm{N\,C^{-1}}. \] The central negative charge attracts a positive test charge to the left: \[ E_M=-\frac{(8.99\times10^9)(6.00\times10^{-9})}{0.150^2}=-2.40\times10^3\,\mathrm{N\,C^{-1}}. \] The right positive charge also pushes a positive test charge to the left: \[ E_R=-\frac{(8.99\times10^9)(3.00\times10^{-9})}{0.150^2}=-1.20\times10^3\,\mathrm{N\,C^{-1}}. \] Thus \[ E_x=1.33\times10^2-2.40\times10^3-1.20\times10^3 =-3.46\times10^3\,\mathrm{N\,C^{-1}}. \] The field magnitude is \(3.46\times10^3\,\mathrm{N\,C^{-1}}\), directed in the \(-x\) direction.

Answer 1. Three-charge field mapping on a line

[15 marks]

c) A \(+2.00\,\mathrm{nC}\) test charge is moved slowly from \(P\) to \(x=+0.600\,\mathrm{m}\). Find the work done by the external agent.

[3 marks]

For a slow move, the external work equals the change in electric potential energy: \[ W_{\mathrm{ext}}=q_0\Delta V=q_0(V_R-V_P). \] At \(x=+0.600\,\mathrm{m}\), the distances from the left, middle, and right charges are \(0.900\,\mathrm{m}\), \(0.600\,\mathrm{m}\), and \(0.300\,\mathrm{m}\), so \[ V_R=(8.99\times10^9)\left(\frac{3.00\times10^{-9}}{0.900}-\frac{6.00\times10^{-9}}{0.600}+\frac{3.00\times10^{-9}}{0.300}\right) =30.0\,\mathrm{V}. \] Therefore \[ W_{\mathrm{ext}}=(2.00\times10^{-9})[30.0-(-119.9)] =3.00\times10^{-7}\,\mathrm{J}. \]

Answer 1. Three-charge field mapping on a line

[15 marks]

d) Find the finite points on the \(x\)-axis where the electric potential is zero.

[3 marks]

The zero-potential condition is \[ \frac{3.00}{|x+0.300|}-\frac{6.00}{|x|}+\frac{3.00}{|x-0.300|}=0. \] By symmetry, any zero-potential points occur as a pair about the origin. For \(0<x<0.300\,\mathrm{m}\), \[ \frac{1}{x+0.300}-\frac{2}{x}+\frac{1}{0.300-x}=0. \] Let \(u=x/0.300\). Then \[ \frac{1}{1+u}-\frac{2}{u}+\frac{1}{1-u}=0. \] Since \(1/(1+u)+1/(1-u)=2/(1-u^2)\), \[ \frac{2}{1-u^2}=\frac{2}{u}. \] Thus \[ u^2+u-1=0,\qquad u=\frac{\sqrt5-1}{2}=0.618. \] So \[ x=0.300(0.618)=0.185\,\mathrm{m}. \] The symmetric point is \(x=-0.185\,\mathrm{m}\). The outer regions give no valid finite solutions.

Answer 1. Three-charge field mapping on a line

[15 marks]

e) Explain why the zero-potential points found in part (d) do not have to be zero-field points.

[2 marks]

Potential is a scalar, so the signed contributions from positive and negative charges can cancel without regard to direction. Electric field is a vector, so the direction of each contribution matters. A zero value of \(V\) at one point does not imply a flat potential there; the field depends on the spatial gradient, \[ \vec E=-\nabla V. \]

Section B: Electromagnetism

Answer 2. RC discharge and inductor safety dissipation

[15 marks]

A safety interlock uses a \(470\,\mu\mathrm{F}\) capacitor initially charged to \(24.0\,\mathrm{V}\). The capacitor discharges through a \(3.30\,\mathrm{k\Omega}\) resistor into a sensor input. When the capacitor voltage falls to \(6.00\,\mathrm{V}\), a separate switch opens an inductor branch that had been carrying \(1.80\,\mathrm{A}\). The inductor has \(L=0.250\,\mathrm{H}\) and then discharges through a \(10.0\,\Omega\) safety resistor.

a) Find the RC time constant of the sensor discharge.

[2 marks]

The RC time constant is \[ \tau_{RC}=RC. \] With \(R=3.30\times10^3\,\Omega\) and \(C=470\times10^{-6}\,\mathrm{F}\), \[ \tau_{RC}=(3.30\times10^3)(470\times10^{-6}) =1.55\,\mathrm{s}. \]

Answer 2. RC discharge and inductor safety dissipation

[15 marks]

b) Calculate the time after which the sensor voltage reaches \(6.00\,\mathrm{V}\).

[3 marks]

For capacitor discharge, \[ V_C(t)=V_0e^{-t/\tau_{RC}}. \] Set \(V_C=6.00\,\mathrm{V}\) and \(V_0=24.0\,\mathrm{V}\): \[ 6.00=24.0e^{-t/1.55}. \] Thus \[ e^{-t/1.55}=0.250,\qquad t=(1.55)\ln4=2.15\,\mathrm{s}. \]

Answer 2. RC discharge and inductor safety dissipation

[15 marks]

c) Find the capacitor energy initially and at the switching voltage, and state where the lost energy goes.

[3 marks]

The energy stored in the capacitor is \[ U_C=\frac12CV^2. \] Initially, \[ U_i=\frac12(470\times10^{-6})(24.0)^2=0.135\,\mathrm{J}. \] At \(6.00\,\mathrm{V}\), \[ U_f=\frac12(470\times10^{-6})(6.00)^2=8.46\times10^{-3}\,\mathrm{J}. \] The energy lost from the capacitor is \[ \Delta U_C=0.135-0.00846=0.127\,\mathrm{J}. \] In this model it is dissipated mainly as thermal energy in the discharge resistor and input circuit.

Answer 2. RC discharge and inductor safety dissipation

[15 marks]

d) Find the magnetic energy stored just before the switch opens and the RL decay time constant afterward.

[3 marks]

The magnetic energy in the inductor just before the switch opens is \[ U_L=\frac12LI^2=\frac12(0.250)(1.80)^2=0.405\,\mathrm{J}. \] Once connected only to the \(10.0\,\Omega\) safety resistor, the RL time constant is \[ \tau_{RL}=\frac{L}{R}=\frac{0.250}{10.0}=2.50\times10^{-2}\,\mathrm{s}. \]

Answer 2. RC discharge and inductor safety dissipation

[15 marks]

e) When the inductor branch opens into the safety resistor, find the initial induced emf magnitude and sign effect, the time for the current to fall to \(0.200\,\mathrm{A}\), and the energy dissipated in that interval.

[4 marks]

Immediately after the switch opens, the current is still \(1.80\,\mathrm{A}\). The resistor drop has magnitude \[ IR=(1.80)(10.0)=18.0\,\mathrm{V}. \] For the inductor, \[ L\frac{dI}{dt}+IR=0, \] so the inductor emf is opposite the chosen current-reference voltage but acts to keep current flowing in its original direction. Its magnitude is \(18.0\,\mathrm{V}\). The current decays as \[ I=I_0e^{-t/\tau_{RL}}. \] For \(I=0.200\,\mathrm{A}\), \[ 0.200=1.80e^{-t/0.0250}, \] so \[ t=(0.0250)\ln9=5.49\times10^{-2}\,\mathrm{s}. \] The remaining magnetic energy is \[ U_f=\frac12(0.250)(0.200)^2=5.00\times10^{-3}\,\mathrm{J}, \] so the energy dissipated in the safety resistor is \[ 0.405-0.00500=0.400\,\mathrm{J}. \]

Section C: Relativity

Answer 3. Fixed-target particle-production threshold

[15 marks]

A proton beam strikes stationary protons in a fixed-target experiment. The proton rest energy is \(m_pc^2=938\,\mathrm{MeV}\). Consider the possible reaction \(p+p\to p+p+X\), where \(X\) is a neutral particle with rest energy \(135\,\mathrm{MeV}\). Use units in which energies are in \(\mathrm{MeV}\) and momenta are in \(\mathrm{MeV}\,c^{-1}\). The relativistic energy-momentum relation is \[ E^2=(pc)^2+(m_pc^2)^2. \] For this fixed-target geometry, the centre-of-mass energy satisfies \[ E_{\mathrm{cm}}^2=2m_p^2c^4+2E m_pc^2, \] where \(E\) is the incident proton's total energy.

a) If the incident proton has kinetic energy \(350\,\mathrm{MeV}\), calculate its total energy and momentum.

[3 marks]

For a beam kinetic energy \(K=350\,\mathrm{MeV}\), the total beam-proton energy is \[ E=K+m_pc^2=350+938=1288\,\mathrm{MeV}. \] The momentum follows from \[ E^2=(pc)^2+(m_pc^2)^2. \] Thus \[ pc=\sqrt{1288^2-938^2}=8.82\times10^2\,\mathrm{MeV}. \] Therefore \[ p=8.82\times10^2\,\mathrm{MeV}\,c^{-1}. \]

Answer 3. Fixed-target particle-production threshold

[15 marks]

b) Find the centre-of-mass energy available for this fixed-target collision.

[3 marks]

For one incident proton and one target proton at rest, the invariant energy satisfies \[ E_{\mathrm{cm}}^2=2m_p^2c^4+2E m_pc^2. \] Using \(E=1288\,\mathrm{MeV}\), \[ E_{\mathrm{cm}}^2=2(938)^2+2(1288)(938). \] Thus \[ E_{\mathrm{cm}}=2.04\times10^3\,\mathrm{MeV}. \]

Answer 3. Fixed-target particle-production threshold

[15 marks]

c) Calculate the minimum incident proton kinetic energy needed to produce particle \(X\).

[4 marks]

At threshold, the final particles have no relative motion in the centre-of-mass frame, so \[ E_{\mathrm{cm,th}}=2m_pc^2+m_Xc^2=2(938)+135=2011\,\mathrm{MeV}. \] The threshold beam total energy is found from \[ (2011)^2=2(938)^2+2E_{\mathrm{th}}(938). \] Hence \[ E_{\mathrm{th}}=\frac{2011^2-2(938)^2}{2(938)} =1.22\times10^3\,\mathrm{MeV}. \] The threshold kinetic energy is \[ K_{\mathrm{th}}=E_{\mathrm{th}}-938=2.80\times10^2\,\mathrm{MeV}. \]

Answer 3. Fixed-target particle-production threshold

[15 marks]

d) For the \(350\,\mathrm{MeV}\) beam, how much centre-of-mass energy is above the \(p+p+X\) threshold? Interpret it.

[3 marks]

For the \(350\,\mathrm{MeV}\) beam, part (b) gave \[ E_{\mathrm{cm}}=2043\,\mathrm{MeV}. \] The threshold rest energy of the final particles is \[ 2011\,\mathrm{MeV}. \] The excess centre-of-mass energy is therefore \[ 2043-2011=32\,\mathrm{MeV}. \] This is available as kinetic energy of the final particles in the centre-of-mass frame, not as additional new rest energy for this specified final state.

Answer 3. Fixed-target particle-production threshold

[15 marks]

e) Explain why the threshold is determined using invariant mass rather than by simply comparing the beam kinetic energy with \(135\,\mathrm{MeV}\).

[2 marks]

At threshold the products are at rest relative to one another in the centre-of-mass frame, so the centre-of-mass energy is just their total rest energy. In the laboratory frame, however, the total momentum is not zero because the target was initially at rest and the beam proton was moving. Some laboratory energy must remain associated with the motion of the centre of mass. This is why a fixed-target threshold kinetic energy is greater than the new particle's rest energy alone.

Section D: Quantum Physics

Answer 4. Photoelectric detection and electron diffraction

[15 marks]

A photoelectric detector uses ultraviolet light of wavelength \(240\,\mathrm{nm}\) on a metal photocathode with work function \(2.10\,\mathrm{eV}\). The emitted electrons then enter an electron-diffraction stage, where the fastest electrons are accelerated through an additional \(150\,\mathrm{V}\) before reaching a thin crystal. For photon calculations, take \[ hc=1240\,\mathrm{eV\,nm}. \]

a) Calculate the photon energy and the maximum initial kinetic energy of the emitted electrons. The photoelectric equation is \[ K_{\max}=\frac{hc}{\lambda}-\phi. \]

[3 marks]

Using \(hc=1240\,\mathrm{eV\,nm}\), \[ E_\gamma=\frac{hc}{\lambda}=\frac{1240}{240}=5.17\,\mathrm{eV}. \] Einstein's photoelectric equation gives \[ K_{\max}=E_\gamma-\phi. \] Therefore \[ K_{\max}=5.17-2.10=3.07\,\mathrm{eV}. \]

Answer 4. Photoelectric detection and electron diffraction

[15 marks]

b) Find the stopping voltage for the detector and state the required polarity.

[2 marks]

The stopping voltage satisfies \[ eV_s=K_{\max}. \] Because \(K_{\max}=3.07\,\mathrm{eV}\), \[ V_s=3.07\,\mathrm{V}. \] The collector must be made negative relative to the photocathode by this magnitude to stop the fastest electrons.

Answer 4. Photoelectric detection and electron diffraction

[15 marks]

c) Calculate the threshold wavelength of the photocathode.

[3 marks]

At threshold, \(K_{\max}=0\), so \[ \frac{hc}{\lambda_0}=\phi. \] Thus \[ \lambda_0=\frac{1240\,\mathrm{eV\,nm}}{2.10\,\mathrm{eV}} =590\,\mathrm{nm}. \] Wavelengths longer than this have photon energy below the work function and do not emit photoelectrons.

Answer 4. Photoelectric detection and electron diffraction

[15 marks]

d) Estimate the de Broglie wavelength of the fastest electrons just before diffraction. For a non-relativistic electron, the momentum is \[ p=\sqrt{2m_eK} \] and the de Broglie wavelength is \[ \lambda=\frac{h}{p}. \]

[5 marks]

The fastest electrons enter the diffraction stage with total kinetic energy approximately \[ K=150\,\mathrm{eV}+3.07\,\mathrm{eV}=153\,\mathrm{eV}. \] Convert to joules: \[ K=(153)(1.60\times10^{-19})=2.45\times10^{-17}\,\mathrm{J}. \] For a non-relativistic electron, \[ p=\sqrt{2m_eK} =\sqrt{2(9.11\times10^{-31})(2.45\times10^{-17})} =6.69\times10^{-24}\,\mathrm{kg\,m\,s^{-1}}. \] The de Broglie wavelength is \[ \lambda=\frac{h}{p}=\frac{6.63\times10^{-34}}{6.69\times10^{-24}} =9.91\times10^{-11}\,\mathrm{m} =0.099\,\mathrm{nm}. \]

Answer 4. Photoelectric detection and electron diffraction

[15 marks]

e) If an electron is localized to \(0.050\,\mathrm{nm}\) as it scatters, estimate the minimum momentum uncertainty and compare it with the momentum from part (d). The Heisenberg uncertainty relation is \[ \Delta x\Delta p\ge\frac{\hbar}{2}. \]

[2 marks]

Using \(\Delta x\Delta p\ge\hbar/2\) with \(\hbar\approx1.05\times10^{-34}\,\mathrm{J\,s}\), \[ \Delta p\ge\frac{1.05\times10^{-34}}{2(0.050\times10^{-9})} =1.05\times10^{-24}\,\mathrm{kg\,m\,s^{-1}}. \] From part (d), \(p=6.69\times10^{-24}\,\mathrm{kg\,m\,s^{-1}}\), so \[ \frac{\Delta p}{p}\gtrsim\frac{1.05}{6.69}=0.16. \] The uncertainty is about \(16\%\) of the electron momentum, large enough to noticeably broaden the diffraction direction.