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Paper 2 Answers

Time2 hours

Marks60

SetSet 3

PaperLevel 1 - Physics Paper 2

Information

Section A: Electromagnetism
Section B: Relativity and Quantum Mechanics

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: Electromagnetism

Answer 1. Charged insulating sheet field cage

[15 marks]

Two square insulating sheets are mounted parallel to one another to make a simple field cage. Each sheet has area \(A=0.120\,\mathrm{m^2}\), and the separation is \(d=3.00\,\mathrm{mm}\). The left sheet has charge \(+Q=8.50\,\mathrm{nC}\) spread uniformly over it, and the right sheet has charge \(-Q\) spread uniformly over it. Treat the sheets as very large compared with their separation, so edge effects may be ignored. The field magnitude from one very large uniformly charged sheet is \[ E=\frac{\sigma}{2\epsilon_0}. \] The field is directed away from a positive sheet and toward a negative sheet.

a) Find the surface charge density and the electric field magnitude inside and outside the sheet pair. State the field direction inside the cage.

[4 marks]

The surface charge density is \[ \sigma=\frac{Q}{A} =\frac{8.50\times10^{-9}}{0.120} =7.08\times10^{-8}\,\mathrm{C\,m^{-2}}. \] A single large sheet gives field magnitude \(\sigma/(2\epsilon_0)\) on each side. Between the sheets, the fields from the positive and negative sheets point from the positive sheet toward the negative sheet, so they add: \[ E=\frac{\sigma}{\epsilon_0} =\frac{7.08\times10^{-8}}{8.85\times10^{-12}} =8.00\times10^3\,\mathrm{N\,C^{-1}}. \] Outside the pair the two sheet fields are equal and opposite, so the approximate external field is zero.

Answer 1. Charged insulating sheet field cage

[15 marks]

b) Calculate the potential difference between the sheets and identify which sheet is at higher potential.

[3 marks]

For a uniform field, the potential difference between the sheets is \[ \Delta V=Ed. \] Thus \[ \Delta V=(8.00\times10^3)(3.00\times10^{-3}) =24.0\,\mathrm{V}. \] The positive sheet is at higher potential because the electric field points from higher potential toward lower potential.

Answer 1. Charged insulating sheet field cage

[15 marks]

c) Find the capacitance per unit area and the total capacitance of the sheet pair. Check your result using \(V=Q/C\).

[4 marks]

The capacitance per unit area is \[ \frac{C}{A}=\frac{\epsilon_0}{d}. \] Numerically, \[ \frac{C}{A}=\frac{8.85\times10^{-12}}{3.00\times10^{-3}} =2.95\times10^{-9}\,\mathrm{F\,m^{-2}}. \] The total capacitance is \[ C=\frac{\epsilon_0 A}{d} =(2.95\times10^{-9})(0.120) =3.54\times10^{-10}\,\mathrm{F}. \] The same voltage follows from \(V=Q/C\): \[ V=\frac{8.50\times10^{-9}}{3.54\times10^{-10}} =24.0\,\mathrm{V}. \]

Answer 1. Charged insulating sheet field cage

[15 marks]

d) Calculate the electrostatic energy stored in the field between the sheets.

[2 marks]

The electrostatic energy is \[ U=\frac12Q\Delta V. \] Therefore \[ U=\frac12(8.50\times10^{-9})(24.0) =1.02\times10^{-7}\,\mathrm{J}. \]

Answer 1. Charged insulating sheet field cage

[15 marks]

e) The sheet pair is isolated and the gap is filled with a dielectric of relative permittivity \(\kappa=2.40\). State the new field, voltage, and capacitance per unit area.

[2 marks]

With fixed free charge, inserting a dielectric changes \(\epsilon_0\) to \(\kappa\epsilon_0\). The field and voltage become \[ E'=\frac{E}{\kappa},\qquad V'=\frac{V}{\kappa}. \] The capacitance per unit area becomes \[ \left(\frac{C}{A}\right)'=\kappa\frac{\epsilon_0}{d}. \] For \(\kappa=2.40\), \[ E'=3.33\times10^3\,\mathrm{N\,C^{-1}}, \qquad V'=10.0\,\mathrm{V}, \] and \[ (C/A)'=2.40(2.95\times10^{-9}) =7.08\times10^{-9}\,\mathrm{F\,m^{-2}}. \]

Answer 2. Crossed-field ion selector

[15 marks]

A beam of singly charged positive ions enters a velocity selector travelling in the \(+\hat{\imath}\) direction. In the selector, \(\vec E=1.80\times10^4\,\hat{\jmath}\,\mathrm{N\,C^{-1}}\) and \(\vec B=0.300\,\hat k\,\mathrm{T}\). Ions that pass undeflected then enter an analyser containing only a uniform magnetic field of magnitude \(0.500\,\mathrm{T}\). Two isotopes have masses \(m_1=3.32\times10^{-26}\,\mathrm{kg}\) and \(m_2=3.49\times10^{-26}\,\mathrm{kg}\).

a) Show that the electric and magnetic forces oppose one another, and find the selected ion speed.

[4 marks]

For a positive ion, the electric force is in the \(+\hat{\jmath}\) direction. The magnetic part is \[ \vec v\times\vec B=\hat{\imath}\times\hat k=-\hat{\jmath}, \] so the magnetic force is in the \(-\hat{\jmath}\) direction. Undeflected motion requires equal magnitudes: \[ qE=qvB. \] Hence \[ v=\frac{E}{B} =\frac{1.80\times10^4}{0.300} =6.00\times10^4\,\mathrm{m\,s^{-1}}. \]

Answer 2. Crossed-field ion selector

[15 marks]

b) State the deflection direction for ions that are slower and faster than the selected speed.

[2 marks]

If an ion is too slow, then \(qvB<qE\), so the electric force is larger and the ion deflects toward \(+\hat{\jmath}\). If it is too fast, then \(qvB>qE\), so it deflects toward \(-\hat{\jmath}\). This direction check is consistent with only \(v=E/B\) passing straight through.

Answer 2. Crossed-field ion selector

[15 marks]

c) Calculate the circular analyser radius for each isotope.

[4 marks]

In the analyser, the magnetic force supplies the centripetal force: \[ qvB_a=\frac{mv^2}{r}. \] Therefore \[ r=\frac{mv}{qB_a}. \] For isotope 1, \[ r_1=\frac{(3.32\times10^{-26})(6.00\times10^4)}{(1.60\times10^{-19})(0.500)} =2.49\times10^{-2}\,\mathrm{m}. \] For isotope 2, \[ r_2=\frac{(3.49\times10^{-26})(6.00\times10^4)}{(1.60\times10^{-19})(0.500)} =2.62\times10^{-2}\,\mathrm{m}. \]

Answer 2. Crossed-field ion selector

[15 marks]

d) Find the difference in analyser radii. If the detector is placed after a semicircle, estimate the separation between the two impact points and identify which isotope lands farther from the entry point.

[3 marks]

The separation of the radii is \[ \Delta r=r_2-r_1 =(2.62-2.49)\times10^{-2} =1.28\times10^{-3}\,\mathrm{m}. \] If the ions make a semicircle before detection, the separation of the impact points is approximately \[ 2\Delta r=2.55\times10^{-3}\,\mathrm{m}. \] The heavier isotope has the larger radius because \(r\propto m\) for fixed \(q\), \(v\), and \(B_a\).

Answer 2. Crossed-field ion selector

[15 marks]

e) For an analyser field in the \(+\hat k\) direction, state the initial bending direction. What changes if the analyser field is reversed?

[2 marks]

With \(\vec v\) in \(+\hat{\imath}\) and analyser field \(\vec B_a\) in \(+\hat k\), \[ \vec v\times\vec B_a=-\hat{\jmath}. \] For positive ions the force is therefore initially in the \(-\hat{\jmath}\) direction. Reversing the analyser field reverses \(\vec v\times\vec B_a\), so the circular path bends toward \(+\hat{\jmath}\) instead. The radius magnitude is unchanged because it depends on \(B_a\), not on the field sign.

Section B: Relativity and Quantum Mechanics

Answer 3. Spacelike platform flashes

[15 marks]

Two camera flashes occur along a straight train platform. In the platform frame \(S\), flash A occurs at \(x_A=0\), \(t_A=0\), and flash B occurs at \(x_B=600\,\mathrm{m}\), \(t_B=1.00\,\mu\mathrm{s}\). Another inertial frame \(S'\) moves along \(+x\) relative to the platform. For a frame \(S'\) moving at speed \(u\) along \(+x\) relative to \(S\), the Lorentz transformation for separations is \[ \Delta t'=\gamma\left(\Delta t-\frac{u\Delta x}{c^2}\right) \] and \[ \Delta x'=\gamma(\Delta x-u\Delta t). \]

a) Calculate \(s^2=c^2\Delta t^2-\Delta x^2\) for the two flashes and classify the separation.

[3 marks]

The separations in \(S\) are \[ \Delta x=600\,\mathrm{m},\qquad \Delta t=1.00\times10^{-6}\,\mathrm{s}. \] The interval is \[ s^2=c^2\Delta t^2-\Delta x^2. \] Since \[ c\Delta t=(3.00\times10^8)(1.00\times10^{-6})=300\,\mathrm{m}, \] \[ s^2=(300)^2-(600)^2 =-2.70\times10^5\,\mathrm{m^2}. \] The interval is spacelike because \(s^2<0\).

Answer 3. Spacelike platform flashes

[15 marks]

b) Find the speed and direction of the inertial frame in which the flashes are simultaneous.

[4 marks]

For simultaneity in \(S'\), \[ \Delta t'=\gamma\left(\Delta t-\frac{u\Delta x}{c^2}\right)=0. \] Thus \[ u=\frac{c^2\Delta t}{\Delta x}. \] Substitute the values: \[ u=\frac{(3.00\times10^8)^2(1.00\times10^{-6})}{600} =1.50\times10^8\,\mathrm{m\,s^{-1}} =0.500c. \] The frame must move in the \(+x\) direction because flash B is at larger \(x\) and later \(t\) in the platform frame.

Answer 3. Spacelike platform flashes

[15 marks]

c) Calculate the transformed spatial separation of the flashes in the frame found in part (b).

[3 marks]

For \(u=0.500c\), \[ \gamma=\frac{1}{\sqrt{1-0.500^2}}=1.155. \] The transformed separation is \[ \Delta x'=\gamma(\Delta x-u\Delta t). \] Here \[ u\Delta t=(1.50\times10^8)(1.00\times10^{-6})=150\,\mathrm{m}. \] Therefore \[ \Delta x'=1.155(600-150) =5.20\times10^2\,\mathrm{m}. \] The positive sign means flash B is still at the larger \(x'\) coordinate.

Answer 3. Spacelike platform flashes

[15 marks]

d) For a frame moving at \(0.800c\) in the \(+x\) direction, find \(\Delta t'\) and state the time order of the flashes.

[3 marks]

For a frame moving at \(0.800c\), \[ \gamma=\frac{1}{\sqrt{1-0.800^2}}=1.67. \] The time separation is \[ \Delta t'=\gamma\left(\Delta t-\frac{u\Delta x}{c^2}\right). \] Now \[ \frac{u\Delta x}{c^2}=\frac{0.800c(600)}{c^2} =\frac{480}{3.00\times10^8} =1.60\,\mu\mathrm{s}. \] Thus \[ \Delta t'=1.67(1.00-1.60)\,\mu\mathrm{s} =-1.00\,\mu\mathrm{s}. \] In this frame, flash B occurs before flash A.

Answer 3. Spacelike platform flashes

[15 marks]

e) Explain why the change of time order in some frames does not violate causality.

[2 marks]

A causal signal would need speed \[ \frac{\Delta x}{\Delta t}=\frac{600}{1.00\times10^{-6}} =6.00\times10^8\,\mathrm{m\,s^{-1}}=2.00c, \] which is greater than the speed of light. The events are spacelike separated, so neither flash can cause the other. Different inertial frames may disagree about their time order without creating a causality contradiction.

Answer 4. Infinite well spectroscopy after widening

[15 marks]

An electron is confined in a one-dimensional infinite square well. The original well width is \(L_0=0.500\,\mathrm{nm}\), but the well is then widened to \(L=0.750\,\mathrm{nm}\). For width \(L\), the stationary-state energies are \[ E_n=\frac{n^2h^2}{8m_eL^2} \] and the normalized wavefunctions are \(\psi_n(x)=\sqrt{2/L}\sin(n\pi x/L)\) for \(0<x<L\).

a) For the widened well, calculate \(E_1\) in joules and electronvolts, then find \(E_2\) and \(E_3\) in electronvolts.

[4 marks]

For the widened well, \[ E_1=\frac{h^2}{8m_eL^2}. \] With \(L=0.750\times10^{-9}\,\mathrm{m}\), \[ E_1=\frac{(6.63\times10^{-34})^2}{8(9.11\times10^{-31})(0.750\times10^{-9})^2} =1.07\times10^{-19}\,\mathrm{J}. \] In electronvolts, \[ E_1=\frac{1.07\times10^{-19}}{1.60\times10^{-19}} =0.669\,\mathrm{eV}. \] The next levels are \[ E_2=4E_1=2.68\,\mathrm{eV}, \qquad E_3=9E_1=6.02\,\mathrm{eV}. \]

Answer 4. Infinite well spectroscopy after widening

[15 marks]

b) Find the wavelength of the photon emitted when an electron falls from \(n=3\) to \(n=2\) in the widened well.

[3 marks]

For the transition \(n=3\to n=2\), \[ \Delta E=E_3-E_2=(9-4)E_1=5E_1. \] Using joules, \[ \Delta E=5(1.07\times10^{-19}) =5.36\times10^{-19}\,\mathrm{J}. \] The emitted photon wavelength is \[ \lambda=\frac{hc}{\Delta E} =\frac{(6.63\times10^{-34})(3.00\times10^8)}{5.36\times10^{-19}} =3.71\times10^{-7}\,\mathrm{m}. \] So \(\lambda=371\,\mathrm{nm}\).

Answer 4. Infinite well spectroscopy after widening

[15 marks]

c) Write \(\psi_2(x)\) and calculate the probability of finding the electron between \(x=L/4\) and \(x=L/2\).

[4 marks]

For \(n=2\), \[ \psi_2(x)=\sqrt{\frac{2}{L}}\sin\left(\frac{2\pi x}{L}\right). \] The probability is \[ P=\int_{L/4}^{L/2}\frac{2}{L}\sin^2\left(\frac{2\pi x}{L}\right)\,dx. \] Using \[ \int \sin^2(kx)\,dx=\frac{x}{2}-\frac{\sin(2kx)}{4k}, \] with \(k=2\pi/L\), \[ P=\left[\frac{x}{L}-\frac{\sin(4\pi x/L)}{4\pi}\right]_{L/4}^{L/2}. \] At \(x=L/2\), \(\sin(2\pi)=0\). At \(x=L/4\), \(\sin(\pi)=0\). Therefore \[ P=\frac12-\frac14=\frac14. \]

Answer 4. Infinite well spectroscopy after widening

[15 marks]

d) By what factor do all energy levels change when the well is widened from \(0.500\,\mathrm{nm}\) to \(0.750\,\mathrm{nm}\)?

[2 marks]

The energies scale as \[ E_n\propto\frac{1}{L^2}. \] Changing from \(L_0=0.500\,\mathrm{nm}\) to \(L=0.750\,\mathrm{nm}\) gives \[ \frac{E_{n,new}}{E_{n,old}}=\left(\frac{L_0}{L}\right)^2 =\left(\frac{0.500}{0.750}\right)^2 =\frac{4}{9}. \] All energy levels and transition energies are reduced by a factor \(4/9\).

Answer 4. Infinite well spectroscopy after widening

[15 marks]

e) State how the photon wavelength for the same transition changes after widening the well, and explain the spectral shift.

[2 marks]

For a fixed pair of quantum numbers, \[ \lambda=\frac{hc}{\Delta E}. \] Since the transition energy is multiplied by \(4/9\), the photon wavelength is multiplied by the reciprocal factor: \[ \frac{\lambda_{new}}{\lambda_{old}}=\frac{9}{4}=2.25. \] The spectrum shifts to longer wavelengths because widening the well makes the energy levels closer together.