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Unit 9: Thermodynamics

1. Topic 9.1 · Say it in contextGas A and gas B are at the same temperature. Each atom of A has four times the mass of an atom of B. Compare their average kinetic energies and their rms speeds.

   The average kinetic energies are equal, because $K_{avg}=\frac{3}{2}{k}_{B}T$ depends only on temperature. Since $\frac{1}{2}m{v}_{rms}^2$ is the same for both, the atoms of A move at half the rms speed of B: four times the mass, $\sqrt{4}=2$ times slower.

   Review: 9.1 Kinetic Theory of Temperature and Pressure

2. Topic 9.1 · Say it in contextA sealed sample of gas is heated. How does its Maxwell-Boltzmann distribution of atomic speeds change?

   The peak moves to a higher speed and the curve spreads out and gets lower, so more atoms have high speeds. The area under the curve stays the same, because the number of atoms has not changed.

   Review: 9.1 Kinetic Theory of Temperature and Pressure

3. Topic 9.2 · Work it outA rigid sealed tank holds an ideal gas at 300 K and 120 kPa. It is heated to 450 K. What is the new pressure?

   The amount of gas and the volume are fixed, so $P/T$ stays constant: $P_2=(120\text{kPa})\frac{450}{300}=180$ kPa. This only works with kelvin temperatures.

   Review: 9.2 The Ideal Gas Law

4. Topic 9.2 · Spot the mistake"A rigid tank of gas warms from 20 degrees C to 40 degrees C, so its pressure doubles." What is wrong?

   Pressure is proportional to absolute temperature, not Celsius. In kelvin that is 293 K to 313 K, a factor of about 1.07, so the pressure rises by about 7 percent. Extrapolate a pressure-temperature graph to zero pressure and you land near $-273$ degrees C, which is why the zero matters.

   Review: 9.2 The Ideal Gas Law

5. Topic 9.3 · Spot the mistakeA metal spoon and a wooden spoon have been in the same drawer all night. The metal one feels colder. "So the metal spoon is at a lower temperature." True?

   False. After a night in thermal contact with the drawer, both are at the same temperature: thermal equilibrium. The metal feels colder because it carries energy away from your warmer hand faster than wood does.

   Review: 9.3 Thermal Energy Transfer and Equilibrium

6. Topic 9.3 · Which one fitsA mug of hot coffee is cooling. Name the thermal process in each: the handle gets warm, warm air rises off the top, and your hand feels warmth beside the mug without touching it.

   The handle warms by conduction through the mug. The rising air is convection, energy carried by moving fluid. The warmth beside the mug without contact is radiation. In all three, energy goes from the hotter system to the colder one.

   Review: 9.3 Thermal Energy Transfer and Equilibrium

7. Topic 9.4 · Work it outA gas absorbs 500 J by heating while it expands at a constant 100 kPa from 2.0 L to 5.0 L. Find the work done on the gas and the change in its internal energy.

   $W=-P\mathrm{\Delta }V=-(1.0\times {10}^5)(3.0\times {10}^{-3})=-300$ J; the gas expanded, so the work done on it is negative. Then $\mathrm{\Delta }U=Q+W=500+(-300)=200$ J.

   Review: 9.4 The First Law of Thermodynamics

8. Topic 9.4 · Spot the mistake"If I know where a gas starts and ends on a PV diagram, I know the work done on it and the heat it took in." What is wrong?

   Only the state quantities are fixed by the endpoints: temperature and internal energy. Work is the area under the path, so two different paths between the same two states do different work. Since $\mathrm{\Delta }U$ is the same on both paths, $Q=\mathrm{\Delta }U-W$ differs too. Work and heat belong to the process, not the state.

   Review: 9.4 The First Law of Thermodynamics

9. Topic 9.5 · Work it outHow much energy does it take to warm 0.50 kg of water from 20 degrees C to 80 degrees C? Use $c=4200\text{J/(kg K)}$.

   $Q=mc\mathrm{\Delta }T=(0.50)(4200)(60)=1.26\times {10}^5$ J. A change of 60 degrees C is a change of 60 K, so either scale works for $\mathrm{\Delta }T$.

   Review: 9.5 Specific Heat and Thermal Conductivity

10. Topic 9.5 · Which one fitsA wall is replaced by one of the same material with twice the area and twice the thickness. The temperature difference across it is unchanged. What happens to the rate of energy flow by conduction?

    It stays the same. $\frac{Q}{\mathrm{\Delta }t}=\frac{kA\mathrm{\Delta }T}{L}$: doubling $A$ doubles the rate and doubling $L$ halves it, so the two changes cancel.

    Review: 9.5 Specific Heat and Thermal Conductivity

11. Topic 9.6 · Spot the mistake"A refrigerator makes the food inside colder and more ordered, so it breaks the second law." What is wrong?

    The inside of the fridge is not an isolated system. Its entropy can decrease because energy is moved out of it. The fridge dumps even more energy into the kitchen, and the total entropy of the isolated whole (food, fridge and room) does not decrease.

    Review: 9.6 Entropy and the Second Law of Thermodynamics

12. Topic 9.6 · Which one fitsA gas goes from state A to state B once by a fast path and once by a slow path. Is the gas's entropy change different for the two paths?

    No. Entropy is a state function, so the gas's entropy change depends only on states A and B. Compare work and heat, which do depend on the path taken.

    Review: 9.6 Entropy and the Second Law of Thermodynamics

Unit 10: Electric Force, Field, and Potential

1. Topic 10.1 · Work it outTwo point charges 0.30 m apart exert a 4.0 N force on each other. One charge is tripled and the distance is doubled to 0.60 m. What is the new force?

   $F\propto \frac{|q_1{q}_2|}{r^2}$, so the force is multiplied by $\frac{3}{2^2}=\frac{3}{4}$: $(4.0)(0.75)=3.0$ N.

   Review: 10.1 Electric Charge and Electric Force

2. Topic 10.1 · Spot the mistakeA $+6\mu \text{C}$ charge sits near a $+2\mu \text{C}$ charge. "The bigger charge pushes harder on the smaller one than the smaller one pushes back." True?

   False. Coulomb's law has the product $|q_1{q}_2|$ in it, the same number for both, and the forces are a Newton's third law pair: equal in magnitude, opposite in direction, along the line joining the charges.

   Review: 10.1 Electric Charge and Electric Force

3. Topic 10.2 · Say it in contextA negatively charged rod is held near, but not touching, a neutral metal sphere on an insulating stand. Is the sphere attracted, repelled, or neither? What is its net charge?

   Attracted. Electrons in the sphere are pushed to the far side, leaving the near side positive. The positive side is closer to the rod, so the attraction beats the repulsion. The net charge is still zero: nothing was transferred, only rearranged.

   Review: 10.2 Conservation of Electric Charge and the Process of Charging

4. Topic 10.2 · Which one fitsYou have a negatively charged rod and a neutral metal sphere. How do you leave the sphere positively charged without the rod ever touching it?

   Hold the rod near the sphere, then ground the sphere with a wire to Earth. The rod pushes electrons down the wire. Remove the ground first, then the rod. The sphere has lost electrons, so it is left positive.

   Review: 10.2 Conservation of Electric Charge and the Process of Charging

5. Topic 10.3 · Spot the mistake"The electric field here points east, so an electron placed here is pushed east." What is wrong?

   The field's direction is the direction of the force on a positive test charge. An electron is negative, so the force on it, $\overrightarrow{F}=q\overrightarrow{E}$ with $q<0$, points west.

   Review: 10.3 Electric Fields

6. Topic 10.3 · Say it in contextA solid metal sphere carries excess charge and has reached electrostatic equilibrium. Where is the excess charge? What is the field inside the metal, and just outside it?

   The excess charge sits on the surface. The field inside the metal is zero. Just outside, the field is perpendicular to the surface, and everywhere outside it is the same as the field of a point charge with the sphere's net charge placed at its center.

   Review: 10.3 Electric Fields

7. Topic 10.4 · Work it outCharges of $+2.0\mu \text{C}$ and $-3.0\mu \text{C}$ are 0.30 m apart. Find the electric potential energy of the system. Use $k=9.0\times {10}^9{\text{N m}}^2/{\text{C}}^2$.

   $U_E=k\frac{q_1{q}_2}{r}=\frac{(9.0\times {10}^9)(2.0\times {10}^{-6})(-3.0\times {10}^{-6})}{0.30}=-0.18$ J. The sign stays in: negative means an external force must do 0.18 J of work to pull them apart to infinity.

   Review: 10.4 Electric Potential Energy

8. Topic 10.5 · Spot the mistakeTwo equal positive charges sit a distance apart. "The electric field at the midpoint is zero, so the electric potential there is zero too." What is wrong?

   The field is zero because the two field vectors point opposite ways and cancel. Potential is a scalar and the two contributions add: each is $\frac{kQ}{r}$ and positive, so the potential at the midpoint is positive, not zero. Zero field does not mean zero potential.

   Review: 10.5 Electric Potential

9. Topic 10.5 · Say it in contextOn a map of equipotential lines drawn every 10 V, the potential falls from left to right and the lines are packed closer together on the left. Where is the field stronger, and which way does it point?

   Stronger on the left, because $|E|=|\frac{\mathrm{\Delta }V}{\mathrm{\Delta }r}|$ and the same 10 V change happens over a shorter distance there. The field is perpendicular to the lines and points toward lower potential, so here it points to the right.

   Review: 10.5 Electric Potential

10. Topic 10.6 · Say it in contextA parallel-plate capacitor stays connected to a battery while its plate separation is doubled. What happens to its capacitance, its charge, and the field between the plates?

    $C=\frac{\kappa {\epsilon }_{0}A}{d}$, so the capacitance halves. The battery holds $\mathrm{\Delta }V$ fixed, so $Q=C\mathrm{\Delta }V$ halves. The field $E=\frac{\mathrm{\Delta }V}{d}$ halves too: the same potential difference over twice the distance.

    Review: 10.6 Capacitors

11. Topic 10.6 · Spot the mistakeA student finds the capacitance of air-filled plates with area $0.020{\text{m}}^2$ and separation 1.0 mm and reports 0.18 F. What went wrong?

    It is off by a factor of about ${10}^9$. $C=\frac{{\epsilon }_{0}A}{d}=\frac{(8.85\times {10}^{-12})(0.020)}{1.0\times {10}^{-3}}=1.8\times {10}^{-10}$ F, which is 0.18 nF or about 180 pF. The digits are right and the power of ten is not: 0.18 comes from leaving off ${\epsilon }_0$'s ${10}^{-12}$ with $d$ still in millimeters, or from a correct 0.18 nF written as farads. A farad is enormous; handheld plates give picofarads.

    Review: 10.6 Capacitors

12. Topic 10.7 · Work it outA proton starts from rest and accelerates through a potential difference of 500 V. How much kinetic energy does it gain?

    The potential energy it loses becomes kinetic energy: $K=|q\mathrm{\Delta }V|=(1.6\times {10}^{-19})(500)=8.0\times {10}^{-17}$ J, which is 500 eV.

    Review: 10.7 Conservation of Electric Energy

13. Topic 10.7 · Spot the mistake"An electron released from rest in an electric field moves toward lower potential, like everything else." What is wrong?

    A positive charge moves toward lower potential. An electron is negative, so $\mathrm{\Delta }{U}_E=q\mathrm{\Delta }V$ is negative when $\mathrm{\Delta }V$ is positive: its potential energy drops as it moves toward HIGHER potential, and that is the way it goes.

    Review: 10.7 Conservation of Electric Energy

Unit 11: Electric Circuits

1. Topic 11.1 · Spot the mistake"A bulb uses up some of the current, so less current comes out of a bulb than goes in." What is wrong?

   Charge is conserved, so the current into the bulb equals the current out. What the bulb takes is energy: each charge loses electric potential energy going through it. Current is the rate charge passes a point, and nothing is used up.

   Review: 11.1 Electric Current

2. Topic 11.2 · Which one fitsA bulb glows in a simple loop with a battery. Someone connects a plain wire from one terminal of the bulb directly to the other. What happens to the bulb?

   It goes out. The wire is a short circuit: a path with essentially no potential difference across it, so there is no potential difference left across the bulb and the charge goes through the wire instead.

   Review: 11.2 Simple Circuits

3. Topic 11.2 · Spot the mistakeThree bulbs and a battery are in one series loop. "Opening a switch between bulb 2 and bulb 3 only turns off bulb 3." True?

   False. An open switch anywhere in a single loop makes it an open circuit, so charge stops flowing everywhere in the loop at once and all three bulbs go out. Current does not flow out of the battery and fill the wire step by step.

   Review: 11.2 Simple Circuits

4. Topic 11.3 · Spot the mistakeA student graphs current on the vertical axis against potential difference on the horizontal axis. The line has slope 0.25 A/V, so she writes $R=0.25\mathrm{\Omega }$. Fix it.

   The slope is $\frac{I}{\mathrm{\Delta }V}$, which is $\frac{1}{R}$, not $R$. So $R=\frac{1}{0.25}=4.0\mathrm{\Omega }$. A straight line through the origin is what tells you the element is ohmic.

   Review: 11.3 Resistance, Resistivity, and Ohm’s Law

5. Topic 11.3 · Work it outA copper wire is replaced by a copper wire twice as long with twice the diameter. How does its resistance compare?

   $R=\frac{\rho \ell }{A}$. Twice the length doubles $R$. Twice the diameter gives four times the cross-sectional area, which divides $R$ by 4. Together: $\frac{2}{4}=\frac{1}{2}$ the resistance.

   Review: 11.3 Resistance, Resistivity, and Ohm’s Law

6. Topic 11.4 · Say it in contextOne bulb is connected to an ideal battery. Then a second identical bulb is added in series. Treating the bulbs as ohmic, how bright is each bulb now compared with the single bulb?

   Dimmer. Each bulb now gets half the battery's potential difference, and $P=\frac{(\mathrm{\Delta }V{)}^2}{R}$, so each bulb runs at one quarter of the original power. Brightness goes with power.

   Review: 11.4 Electric Power

7. Topic 11.4 · Work it outA heater on a 12 V source draws 3.0 A. What is its power, and what is its resistance?

   $P=I\mathrm{\Delta }V=(3.0)(12)=36$ W. $R=\frac{\mathrm{\Delta }V}{I}=\frac{12}{3.0}=4.0\mathrm{\Omega }$. Check: $I^{2}R=(9.0)(4.0)=36$ W.

   Review: 11.4 Electric Power

8. Topic 11.5 · Spot the mistakeOne bulb glows on an ideal battery. "If I add a second identical bulb in parallel, the battery's current gets shared, so the first bulb dims." What is wrong?

   An ideal battery holds the potential difference fixed, not the current. The first bulb still has the full battery voltage across it, so it is just as bright. The battery now supplies twice the current, one share for each branch.

   Review: 11.5 Compound Direct Current (DC) Circuits

9. Topic 11.5 · Work it outA 6.0 $\mathrm{\Omega }$ and a 3.0 $\mathrm{\Omega }$ resistor are in parallel, and that pair is in series with a 4.0 $\mathrm{\Omega }$ resistor and an ideal 12 V battery. Find the battery current and the current in the 3.0 $\mathrm{\Omega }$ resistor.

   Parallel pair: $\frac{1}{R}=\frac{1}{6}+\frac{1}{3}$, so $R=2.0\mathrm{\Omega }$. Total $6.0\mathrm{\Omega }$, so $I=\frac{12}{6.0}=2.0$ A. The pair has $(2.0)(2.0)=4.0$ V across it, so the 3.0 $\mathrm{\Omega }$ resistor carries $\frac{4.0}{3.0}\approx 1.3$ A.

   Review: 11.5 Compound Direct Current (DC) Circuits

10. Topic 11.6 · Work it outAn ideal 12 V battery drives current through a 2.0 $\mathrm{\Omega }$ and a 4.0 $\mathrm{\Omega }$ resistor in one series loop. Find the potential difference across each resistor.

    $I=\frac{12}{2.0+4.0}=2.0$ A, so the drops are $(2.0)(2.0)=4.0$ V and $(2.0)(4.0)=8.0$ V. Around the loop, $+12-4.0-8.0=0$, as the loop rule requires.

    Review: 11.6 Kirchhoff’s Loop Rule

11. Topic 11.7 · Work it outAt a junction, 2.0 A comes in on one wire and 0.5 A on another. A third wire carries 4.0 A out. What is the current in the fourth wire, and which way?

    Current in must equal current out. In: $2.0+0.5=2.5$ A. Out: 4.0 A. The fourth wire must bring in the missing 1.5 A, so it carries 1.5 A INTO the junction. Charge cannot pile up at a point.

    Review: 11.7 Kirchhoff’s Junction Rule

12. Topic 11.8 · Say it in contextAn uncharged capacitor, a resistor and an ideal battery are in series. The switch closes. Describe the capacitor's potential difference and the current just after closing, and after a long time.

    Just after closing, the uncharged capacitor acts like a wire: 0 V across it, and the current is $\frac{ℰ}{R}$, its largest value. After a long time the capacitor is fully charged: it has the full battery voltage across it and the current is zero.

    Review: 11.8 Resistor-Capacitor (RC) Circuits

13. Topic 11.8 · Work it outA 2.0 k$\mathrm{\Omega }$ resistor charges two 6.0 $\mu$F capacitors connected in series. Find the equivalent capacitance and the time constant.

    In series, $\frac{1}{C}=\frac{1}{6.0}+\frac{1}{6.0}$, so $C_{eq}=3.0\mu \text{F}$, less than either one. $\tau =RC=(2.0\times {10}^3)(3.0\times {10}^{-6})=6.0\times {10}^{-3}$ s. After 6.0 ms the capacitors hold about 63 percent of their final charge.

    Review: 11.8 Resistor-Capacitor (RC) Circuits

Unit 12: Magnetism and Electromagnetism

1. Topic 12.1 · Spot the mistake"Break a bar magnet in half and you get a north piece and a south piece." What happens instead?

   Each half is a complete magnet with its own north and south pole. Magnetism comes from dipoles, from the motion of electrons inside the material, and nobody has ever found a single pole on its own.

   Review: 12.1 Magnetic Fields

2. Topic 12.2 · Work it outA proton moves at $2.0\times {10}^5$ m/s perpendicular to a 0.50 T magnetic field. What is the magnitude of the magnetic force? What if it moved parallel to the field?

   Perpendicular: $F_B=qvB\sin {90}^{\circ }=(1.6\times {10}^{-19})(2.0\times {10}^5)(0.50)=1.6\times {10}^{-14}$ N. Parallel, $\sin 0^{\circ }=0$, so there is no magnetic force at all.

   Review: 12.2 Magnetism and Moving Charges

3. Topic 12.2 · Spot the mistake"A strong magnetic field will speed up an electron that moves through it." What is wrong?

   The magnetic force is always perpendicular to the velocity. It changes the direction of motion but never pushes along it, so it does no work and the speed stays the same. A charge at rest feels no magnetic force at all.

   Review: 12.2 Magnetism and Moving Charges

4. Topic 12.3 · Work it outA long straight wire carries 10 A. What is the magnetic field 2.0 cm from it? Use ${\mu }_0=4\pi \times {10}^{-7}\text{T m/A}$.

   $B=\frac{{\mu }_{0}I}{2\pi r}=\frac{(4\pi \times {10}^{-7})(10)}{2\pi (0.020)}=1.0\times {10}^{-4}$ T. The field circles the wire, tangent to circles centered on it, never pointing toward or away from it.

   Review: 12.3 Magnetism and Current-Carrying Wires

5. Topic 12.3 · Which one fitsTwo long parallel wires carry current in the same direction. Do they attract or repel?

   They attract. Each wire sits in the other's magnetic field, and the right-hand rule gives a force on each one pointing toward the other. Currents in opposite directions repel.

   Review: 12.3 Magnetism and Current-Carrying Wires

6. Topic 12.4 · Spot the mistakeA wire loop sits still in a strong, steady, uniform magnetic field, with the field straight through it. "Lots of flux, so lots of induced emf." What is wrong?

   The emf depends on how fast the flux CHANGES: $|ℰ|=|\frac{\mathrm{\Delta }{\mathrm{\Phi }}_B}{\mathrm{\Delta }t}|$. Nothing here changes, so the emf is zero no matter how large the flux is. Move the loop out of the field, turn it, or change $B$, and you get an emf.

   Review: 12.4 Electromagnetic Induction and Faraday’s Law

7. Topic 12.4 · Say it in contextA bar magnet, north pole first, moves toward a wire loop. You stand behind the magnet looking at the loop. Which way does the induced current circulate, as you see it?

   Counterclockwise. The field from the north pole points away from you through the loop, and that flux is increasing. The induced current makes a field toward you to oppose the CHANGE, and a counterclockwise current gives that. It opposes the change in flux, not the magnet's field itself.

   Review: 12.4 Electromagnetic Induction and Faraday’s Law

Unit 13: Geometric Optics

1. Topic 13.1 · Spot the mistakeA ray hits a flat mirror at 30 degrees measured from the mirror's surface. "So it reflects at 30 degrees." What angle of reflection belongs on the diagram?

   Angles are measured from the normal, the line perpendicular to the surface. The ray is 30 degrees from the surface, so it is 60 degrees from the normal, and it reflects at 60 degrees from the normal on the other side.

   Review: 13.1 Reflection

2. Topic 13.2 · Work it outAn object stands 30 cm in front of a concave mirror with focal length 10 cm. Where is the image, and describe it.

   $\frac{1}{s_i}=\frac{1}{10}-\frac{1}{30}=\frac{2}{30}$, so $s_i=15$ cm in front of the mirror. It is real, inverted and reduced: $|M|=\frac{15}{30}=0.5$.

   Review: 13.2 Images Formed by Mirrors

3. Topic 13.2 · Which one fitsPlane, convex or concave: which mirror always gives an upright, reduced, virtual image, wherever the object is?

   Convex. A plane mirror's image is always upright and virtual but the same size. A concave mirror can give a real inverted image or an upright enlarged virtual one, depending on where the object sits relative to the focal point.

   Review: 13.2 Images Formed by Mirrors

4. Topic 13.3 · Work it outLight in water ($n=1.33$) hits the surface at 30 degrees from the normal. At what angle does it leave into air? What is the critical angle?

   $1.33\sin {30}^{\circ }=(1.00)\sin {\theta }_2$, so $\sin {\theta }_2=0.665$ and ${\theta }_2\approx {42}^{\circ }$, bent away from the normal. Critical angle: ${\sin }^{-1}(\frac{1.00}{1.33})\approx {49}^{\circ }$. Beyond that, no light gets out.

   Review: 13.3 Refraction

5. Topic 13.3 · Spot the mistake"At a steep enough angle, light going from air into glass is totally internally reflected." What is wrong?

   Total internal reflection needs light heading into a medium with a LOWER index of refraction, like glass into air. Going from air into glass the ray bends toward the normal, and there is always an angle it can take, so some light always gets in.

   Review: 13.3 Refraction

6. Topic 13.4 · Spot the mistakeA converging lens forms a sharp image of a candle on a screen. You cover the top half of the lens with card. "Now only half the candle shows on the screen." True?

   False. Light from every point on the candle passes through every part of the lens, so the lower half alone still forms the whole image. The image is complete, just dimmer, because half as much light gets through.

   Review: 13.4 Images Formed by Lenses

7. Topic 13.4 · Work it outAn object is 10 cm from a converging lens with focal length 20 cm. Find the image distance and describe the image.

   $\frac{1}{s_i}=\frac{1}{20}-\frac{1}{10}=-\frac{1}{20}$, so $s_i=-20$ cm. The negative sign means the image is on the same side as the object: virtual and upright, with $|M|=\frac{20}{10}=2$. That is a magnifying glass.

   Review: 13.4 Images Formed by Lenses

Unit 14: Waves, Sound, and Physical Optics

1. Topic 14.1 · Spot the mistakeA pulse travels down a long rope. "The piece of rope at the start rides along with the pulse to the far end." What is wrong?

   A wave carries energy, not matter. Each piece of rope moves up and down, perpendicular to the direction the pulse travels, and returns to where it was. The disturbance moves along; the rope does not.

   Review: 14.1 Properties of Wave Pulses and Waves

2. Topic 14.1 · Work it outA string is under 40 N of tension and has a mass per length of 0.010 kg/m. What is the wave speed? What happens to it if the tension is quadrupled?

   $v=\sqrt{\frac{F_T}{m/\ell }}=\sqrt{\frac{40}{0.010}}=\sqrt{4000}\approx 63$ m/s. Four times the tension doubles the speed, to about 126 m/s.

   Review: 14.1 Properties of Wave Pulses and Waves

3. Topic 14.2 · Work it outA 680 Hz sound travels through air at 340 m/s. What are its wavelength and its period?

   $\lambda =\frac{v}{f}=\frac{340}{680}=0.50$ m. $T=\frac{1}{f}=\frac{1}{680}\approx 1.5\times {10}^{-3}$ s.

   Review: 14.2 Periodic Waves

4. Topic 14.2 · Spot the mistake"If I shake the end of a rope faster, the waves travel down it faster." What is wrong?

   The speed is set by the rope: its tension and mass per length. Shaking faster raises the frequency, and since $v=f\lambda$ stays the same, the wavelength gets shorter.

   Review: 14.2 Periodic Waves

5. Topic 14.3 · Say it in contextAn upward pulse on a light string reaches a knot joining it to a much heavier string. Describe the reflected pulse and the transmitted pulse.

   Waves travel slower on the heavier string, so the reflected pulse comes back inverted (downward). The transmitted pulse continues upright into the heavy string. Whenever a wave changes medium, its frequency stays the same.

   Review: 14.3 Boundary Behavior of Waves and Polarization

6. Topic 14.4 · Which one fitsPut these in order from longest wavelength to shortest: X-rays, radio waves, visible light, ultraviolet, microwaves.

   Radio waves, microwaves, visible light, ultraviolet, X-rays. Infrared would fit between microwaves and visible, and gamma rays after X-rays. Within visible light, red has the longest wavelength and violet the shortest.

   Review: 14.4 Electromagnetic Waves

7. Topic 14.5 · Spot the mistakeAn ambulance drives toward you at a steady speed, passes, and drives away. "The pitch keeps rising as it gets closer." What do you actually hear?

   As it approaches at a steady speed you hear a pitch higher than the siren's rest frequency, and it does not rise. As it passes, the pitch drops, and while it moves away you hear a pitch lower than the rest frequency. The shift depends on how fast the source moves toward or away from you, not on how close it is, so the drop comes as it goes by.

   Review: 14.5 The Doppler Effect

8. Topic 14.6 · Work it outTwo tuning forks, 440 Hz and 444 Hz, are struck together. What beat frequency do you hear?

   $|f_1-f_2|=|440-444|=4$ Hz: the loudness swells and fades four times a second. The two waves drift in and out of phase, which is what makes the beats.

   Review: 14.6 Wave Interference and Standing Waves

9. Topic 14.6 · Work it outA pipe closed at one end is 0.85 m long, and sound travels at 340 m/s. What are its two lowest resonant frequencies?

   Closed at one end means a node there and an antinode at the open end, so the fundamental fits a quarter wavelength: $\lambda =4(0.85)=3.4$ m and $f=\frac{340}{3.4}=100$ Hz. Only odd harmonics fit, so the next is $3\times 100=300$ Hz, not 200 Hz.

   Review: 14.6 Wave Interference and Standing Waves

10. Topic 14.7 · Spot the mistakeLaser light passes through a single slit. "Make the slit narrower and the central bright band gets narrower too." What is wrong?

    It gets wider. The first dark band sits where $a\frac{y_{min}}{L}\approx \lambda$, so $y_{min}$ is proportional to $\frac{1}{a}$. A narrower slit spreads the light more. Diffraction is strongest when the opening is comparable to the wavelength.

    Review: 14.7 Diffraction

11. Topic 14.7 · Which one fitsYou can hear someone talking around the edge of an open doorway but you cannot see them. Why the difference?

    Diffraction is pronounced when the opening is comparable to the wavelength. Speech has wavelengths around a meter, about the size of the doorway, so sound spreads around the edge. Visible light has wavelengths near 500 nm, far smaller, so it barely bends.

    Review: 14.7 Diffraction

12. Topic 14.8 · Work it outLight of wavelength 600 nm passes through two slits 0.20 mm apart onto a screen 2.0 m away. How far apart are neighboring bright fringes?

    $y=\frac{m\lambda L}{d}$, so the spacing is $\frac{(600\times {10}^{-9})(2.0)}{0.20\times {10}^{-3}}=6.0\times {10}^{-3}$ m, or 6.0 mm. The angles are tiny, so the small-angle form is fine.

    Review: 14.8 Double-Slit Interference and Diffraction Gratings

13. Topic 14.8 · Say it in contextIn a double-slit pattern, what is the path length difference from the two slits to the second-order bright fringe (m = 2, counting the central maximum as m = 0)? To the dark band between the first-order and second-order bright fringes?

    At the second bright fringe the waves arrive in step, so the path difference is $2\lambda$. At the dark band between the first and second bright fringes it is $1.5\lambda$: the waves arrive half a wavelength out of step and cancel.

    Review: 14.8 Double-Slit Interference and Diffraction Gratings

14. Topic 14.9 · Work it outAn antireflection coating with $n=1.38$ goes on glass with $n=1.50$. How thick should it be to cancel reflected 550 nm light at normal incidence?

    Both reflections, air to coating and coating to glass, go into a higher index, so both shift by 180 degrees and those shifts cancel. The extra path must be half a wavelength in the coating: $2t=\frac{\lambda }{2n}$, so $t=\frac{550}{4(1.38)}\approx 100$ nm.

    Review: 14.9 Thin-Film Interference

15. Topic 14.9 · Say it in contextLight hits a soap film in air head-on. Is there a 180 degree phase change at the front surface? At the back surface?

    At the front, light in air reflects off soap, which has a higher index, so yes: a 180 degree change. At the back, light inside the soap reflects off air, which has a lower index, so no phase change. That difference decides which film thicknesses look bright.

    Review: 14.9 Thin-Film Interference

Unit 15: Modern Physics

1. Topic 15.1 · Work it outWhat is the energy of a photon of 500 nm light, in electron volts? Use $hc=1240\text{eV nm}$.

   $E=hf=\frac{hc}{\lambda }=\frac{1240}{500}\approx 2.5$ eV (2.48 eV). Shorter wavelength means higher frequency and more energy per photon.

   Review: 15.1 Quantum Theory and Wave-Particle Duality

2. Topic 15.1 · Which one fitsAn electron and a proton move at the same speed. Which has the longer de Broglie wavelength?

   The electron. $\lambda =\frac{h}{p}$, and at the same speed the electron has far less momentum because its mass is about 1800 times smaller, so its wavelength is about 1800 times longer.

   Review: 15.1 Quantum Theory and Wave-Particle Duality

3. Topic 15.2 · Work it outHow many protons and how many neutrons are in a nucleus of ${}_{92}^{235}\text{U}$? What would make a different isotope of uranium?

   92 protons, the lower number, and $235-92=143$ neutrons. Another isotope of uranium still has 92 protons but a different number of neutrons; changing the proton number makes a different element.

   Review: 15.2 The Bohr Model of Atomic Structure

4. Topic 15.3 · Work it outA single-electron atom has levels at $-13.6$ eV, $-3.40$ eV and $-1.51$ eV. Find the energy and wavelength of the photon emitted when the electron drops from $-1.51$ eV to $-3.40$ eV. Use $hc=1240\text{eV nm}$.

   $E=-1.51-(-3.40)=1.89$ eV. $\lambda =\frac{1240}{1.89}\approx 656$ nm, red light. One transition gives one photon of one wavelength.

   Review: 15.3 Emission and Absorption Spectra

5. Topic 15.3 · Spot the mistakeA single-electron atom has levels at $-13.6$, $-3.40$ and $-1.51$ eV. In its ground state it meets a 5.0 eV photon. "It absorbs the photon and moves partway up." What is wrong?

   An atom absorbs only a photon whose energy matches the gap between two of its levels. $-13.6+5.0=-8.6$ eV is not a level, so the photon is not absorbed and passes by. From the ground state, 10.2 eV, 12.09 eV or 13.6 eV and above would work.

   Review: 15.3 Emission and Absorption Spectra

6. Topic 15.4 · Work it outTwo stars of the same size behave as blackbodies. Star A's spectrum peaks at 400 nm and star B's at 800 nm. Which is hotter, by what factor, and how do their powers compare?

   ${\lambda }_{max}=\frac{b}{T}$, so half the peak wavelength means twice the temperature: A is twice as hot. With the same surface area, $P=A\sigma T^4$ gives A $2^4=16$ times the power of B.

   Review: 15.4 Blackbody Radiation

7. Topic 15.5 · Spot the mistake"Make the light brighter and the photoelectrons come off faster." What actually changes?

   Brighter light at the same frequency means more photons, so more electrons come off each second. Each photon still has $hf$, so $K_{max}=hf-\varphi$ does not change. Below the threshold frequency, no electrons come off at all, however bright the light.

   Review: 15.5 The Photoelectric Effect

8. Topic 15.5 · Work it outPhotons of 4.0 eV hit a metal with a work function of 2.3 eV. What is the maximum kinetic energy of the ejected electrons, and what stopping potential would bring the current to zero?

   $K_{max}=hf-\varphi =4.0-2.3=1.7$ eV. An electron with 1.7 eV is stopped by a potential difference of 1.7 V, so the current drops to zero at a stopping potential of 1.7 V.

   Review: 15.5 The Photoelectric Effect

9. Topic 15.6 · Work it outA photon scatters off a free electron and comes straight back ($\theta ={180}^{\circ }$). By how much does its wavelength change? Use $\frac{h}{m_{e}c}=2.43\times {10}^{-12}$ m.

   $\mathrm{\Delta }\lambda =\frac{h}{m_{e}c}(1-\cos {180}^{\circ })=(2.43\times {10}^{-12})(2)\approx 4.86\times {10}^{-12}$ m longer. This is the largest possible shift: the photon gave the electron as much energy as it can.

   Review: 15.6 Compton Scattering

10. Topic 15.7 · Work it outA radioactive sample has a half-life of 8.0 days. You start with 160 g. How much is left after 24 days?

    24 days is $\frac{24}{8.0}=3$ half-lives: $160\to 80\to 40\to 20$ g. You cannot say which nucleus decays when, only how many on average.

    Review: 15.7 Fission, Fusion, and Nuclear Decay

11. Topic 15.7 · Say it in contextIn a fusion reaction the products have $3.0\times {10}^{-29}$ kg less mass than the reactants. How much energy is released, and where does it go?

    $E=\mathrm{\Delta }m{c}^2=(3.0\times {10}^{-29})(3.00\times {10}^8{)}^2=2.7\times {10}^{-12}$ J. It appears as kinetic energy of the products or as photons. Nucleon number is still conserved; what goes missing is mass.

    Review: 15.7 Fission, Fusion, and Nuclear Decay

12. Topic 15.8 · Work it outUranium-238 ($Z=92$) undergoes alpha decay. What are the atomic number and mass number of the nucleus left behind?

    An alpha particle is a helium-4 nucleus: 2 protons and 2 neutrons. The new nucleus has $Z=92-2=90$ and $A=238-4=234$, which is thorium-234. Nucleon number and charge both balance.

    Review: 15.8 Types of Radioactive Decay

13. Topic 15.8 · Spot the mistake"In beta-minus decay, one of the atom's orbiting electrons gets thrown out, so the atomic number stays the same." What is wrong?

    The electron comes from the nucleus. A neutron turns into a proton, emitting an electron and an antineutrino. The atomic number goes up by 1 and the mass number stays the same: carbon-14 becomes nitrogen-14.

    Review: 15.8 Types of Radioactive Decay