AP Physics 2 Modern Physics — Worked Answer Explanations

Unit 6 · 18% of the AP exam · 12 questions explained

1. Question 1 · Easy

   In the photoelectric effect, which property of light determines the maximum kinetic energy of the ejected electrons?

   • A
     The intensity (brightness) of the light

     Why not A: Intensity determines the number of photons (and thus the number of ejected electrons), not the energy of each electron.
   • B
     The frequency (or wavelength) of the lightCorrect
   • C
     The angle of incidence of the light on the metal

     Why not C: The photoelectric effect does not depend on the angle of incidence of the light.
   • D
     The total duration of illumination

     Why not D: Duration affects how many electrons are ejected over time, not the energy per electron.
   Explanation

   Einstein's explanation of the photoelectric effect (Nobel Prize 1921): each photon carries energy $E=hf$ (where $h$ is Planck's constant and $f$ is frequency). When a photon strikes a metal surface, it gives all its energy to one electron. The maximum kinetic energy of the ejected electron is:
   $$K_{\text{max}}=hf-\varphi$$
   where $\varphi$ is the work function (minimum energy to eject an electron). Higher frequency → more energetic photons → faster electrons. Higher intensity → more photons → more electrons ejected, but no change in their individual kinetic energy.

   Key takeaway

   Photoelectric effect: $K_{\text{max}} = hf - \phi$; frequency controls electron energy, intensity controls electron count.

2. Question 2 · Easy

   In the Bohr model of hydrogen, an electron transitions from energy level $n=3$ to $n=1$. This transition results in:

   • A
     The absorption of a photon with energy equal to $E_3-E_1$

     Why not A: Absorption occurs when an electron moves to a higher energy level ($n$ increases); here the electron drops from $n=3$ to $n=1$.
   • B
     The emission of a photon with energy equal to $E_3-E_1$Correct
   • C
     The emission of a photon with energy equal to $E_1-E_3$

     Why not C: $E_1-E_3$ is negative (since $E_3>E_1$ as both are negative but $E_3$ is less negative); photon energy must be positive.
   • D
     No photon is emitted; the electron just loses energy.

     Why not D: Energy must be conserved; the energy lost by the electron must be carried away by a photon.
   Explanation

   In the Bohr model, electrons occupy discrete energy levels. When an electron drops from a higher level $n_i$ to a lower level $n_f$, it emits a photon with energy:
   $$E_{\text{photon}}=E_i-E_f=E_3-E_1$$

   For hydrogen: $E_n=-13.6/n^2\mathrm{eV}$, so:

   • $E_3=-13.6/9\approx -1.51\mathrm{eV}$
   • $E_1=-13.6\mathrm{eV}$
   • $E_{\text{photon}}=(-1.51)-(-13.6)=12.09\mathrm{eV}$ (UV — Lyman series)

   Upward transitions (increasing $n$) require photon absorption; downward transitions emit photons.

   Key takeaway

   Electron dropping to lower $n$: emits photon with $E = E_i - E_f > 0$. Rising to higher $n$: absorbs photon.

3. Question 3 · Easy

   The de Broglie wavelength of a particle with momentum $p$ is given by:

   • A
     $\lambda =hp$

     Why not A: Wavelength decreases with increasing momentum; the formula has $p$ in the denominator, not the numerator.
   • B
     $\lambda =h/p$Correct
   • C
     $\lambda =p/h$

     Why not C: This inverts the de Broglie relation; the wavelength would increase with momentum, which contradicts observation.
   • D
     $\lambda =h/E$ where $E$ is kinetic energy

     Why not D: $h/E$ is not the de Broglie relation; the formula uses momentum $p$, not energy $E$. (For photons, $E=hf=hc/\lambda$, but de Broglie uses $p$.)
   Explanation

   de Broglie's hypothesis (1924): every particle has an associated wavelength:
   $$\lambda =\frac{h}{p}=\frac{h}{mv}$$
   where $h=6.626\times 10^{-34}\mathrm{J}\cdot \mathrm{s}$ is Planck's constant and $p=mv$ is momentum. This is the wave-particle duality for matter. The faster or heavier a particle, the shorter its de Broglie wavelength. For macroscopic objects, $\lambda$ is immeasurably small; for electrons, it is comparable to atomic spacing — hence electron diffraction.

   Key takeaway

   de Broglie wavelength: $\lambda = h/p$; all matter has wave properties, observable when $\lambda$ is comparable to the relevant scale.

4. Question 4 · Easy

   A photon has a wavelength of $\lambda =400\mathrm{nm}$. What is its energy in electron-volts? ($h=6.63\times 10^{-34}\mathrm{J}\cdot \mathrm{s}$, $c=3\times 10^{8}m/s$, $1\mathrm{eV}=1.6\times 10^{-19}\mathrm{J}$)

   • A
     $E\approx 1.24\mathrm{eV}$

     Why not A: This would apply for $\lambda =1000\mathrm{nm}$; using the shortcut $E=1240\mathrm{eV}\cdot \mathrm{nm}/\lambda$ gives $1240/1000=1.24\mathrm{eV}$ for infrared, not UV.
   • B
     $E\approx 3.1\mathrm{eV}$Correct
   • C
     $E\approx 6.2\mathrm{eV}$

     Why not C: This doubles the correct answer, possibly from using $\lambda =200\mathrm{nm}$ or dividing by $0.5$ somewhere.
   • D
     $E\approx 0.31\mathrm{eV}$

     Why not D: This is ten times smaller than the correct answer, likely from a unit error ($\lambda$ in mm instead of nm, or missing a factor of $10^9$).
   Explanation

   Photon energy: $E=hf=hc/\lambda$

   Direct calculation:
   $$E=\frac{(6.63\times 10^{-34})(3\times 10^8)}{400\times 10^{-9}}=\frac{1.989\times 10^{-25}}{4\times 10^{-7}}=4.97\times 10^{-19}\mathrm{J}$$

   Convert to eV:
   $$E=\frac{4.97\times 10^{-19}}{1.6\times 10^{-19}}\approx 3.1\mathrm{eV}$$

   Shortcut: $E=1240\mathrm{eV}\cdot \mathrm{nm}/\lambda =1240/400=3.1\mathrm{eV}$ ✓

   This is a violet photon (visible spectrum: $400-700\mathrm{nm}$).

   Key takeaway

   Photon energy shortcut: $E[\mathrm{eV}] = 1240/\lambda[\mathrm{nm}]$; memorize $hc = 1240\,\mathrm{eV\cdot nm}$.

5. Question 5 · Medium

   Light of frequency $f=1.5\times 10^{15}\mathrm{Hz}$ illuminates a metal with work function $\varphi =3.6\mathrm{eV}$. What is the maximum kinetic energy of the emitted electrons in eV? ($h=4.14\times 10^{-15}\mathrm{eV}\cdot \mathrm{s}$)

   • A
     $K_{\text{max}}=2.61\mathrm{eV}$Correct
   • B
     $K_{\text{max}}=6.21\mathrm{eV}$

     Why not B: This is the photon energy $hf$ alone, without subtracting the work function.
   • C
     $K_{\text{max}}=3.6\mathrm{eV}$

     Why not C: This is just the work function $\varphi$; the photoelectric equation requires $hf-\varphi$, not $\varphi$ alone.
   • D
     $K_{\text{max}}=9.81\mathrm{eV}$

     Why not D: This adds $hf+\varphi$ instead of subtracting; kinetic energy equals photon energy minus work function.
   Explanation

   Photoelectric equation: $K_{\text{max}}=hf-\varphi$

   Photon energy:
   $$hf=(4.14\times 10^{-15}\mathrm{eV}\cdot \mathrm{s})(1.5\times 10^{15}\mathrm{Hz})=6.21\mathrm{eV}$$

   Maximum kinetic energy:
   $$K_{\text{max}}=6.21-3.6=2.61\mathrm{eV}$$

   This electron has enough energy to escape ($hf>\varphi$), with $2.61\mathrm{eV}$ remaining as kinetic energy. The stopping potential needed to arrest this electron would be $V_s=K_{\text{max}}/e=2.61\mathrm{V}$.

   Key takeaway

   Photoelectric: $K_{\text{max}} = hf - \phi$; compute $hf$ first, subtract work function $\phi$; if $K_{\text{max}} < 0$, no emission.

6. Question 6 · Medium

   In nuclear physics, a nucleus with mass number $A=235$ and atomic number $Z=92$ (uranium-235) undergoes alpha decay. What are the mass number and atomic number of the daughter nucleus?

   • A
     $A=231$, $Z=90$Correct
   • B
     $A=233$, $Z=90$

     Why not B: An alpha particle has $A=4$; the daughter mass number must decrease by 4, giving $235-4=231$, not 233.
   • C
     $A=231$, $Z=91$

     Why not C: An alpha particle has $Z=2$; the daughter atomic number decreases by 2, giving $92-2=90$, not 91.
   • D
     $A=235$, $Z=90$

     Why not D: The mass number must also decrease (alpha carries 4 nucleons); only beta decay leaves $A$ unchanged.
   Explanation

   Alpha decay: the parent nucleus emits an alpha particle ${}_2^4\mathrm{He}$ (2 protons, 2 neutrons).

   Conservation laws require:

   • Mass number: $A_{\text{daughter}}=A_{\text{parent}}-4=235-4=231$
   • Atomic number: $Z_{\text{daughter}}=Z_{\text{parent}}-2=92-2=90$

   The daughter is thorium-231 (${}_{90}^{231}\mathrm{Th}$):
   $${}_{92}^{235}\mathrm{U}\to {}_{90}^{231}\mathrm{Th}+{}_2^4\mathrm{He}$$

   Both baryon number (mass number) and charge (atomic number) are conserved.

   Key takeaway

   Alpha decay: daughter $A$ decreases by 4, $Z$ decreases by 2 (alpha = ${}^4_2\mathrm{He}$). Conserve both $A$ and $Z$.

7. Question 7 · Medium

   The half-life of a radioactive isotope is $T_{1/2}=5\mathrm{days}$. Starting with $N_0=1600$ atoms, how many remain after $20\mathrm{days}$?

   • A
     $N=400$ atoms

     Why not A: After 20 days (4 half-lives), the count is $N_0\times (1/2{)}^4=1600/16=100$, not 400.
   • B
     $N=200$ atoms

     Why not B: This corresponds to 3 half-lives ($1600/2^3=200$), but 20 days spans 4 half-lives.
   • C
     $N=100$ atomsCorrect
   • D
     $N=0$ atoms

     Why not D: Exponential decay never reaches exactly zero; after each half-life half the remaining atoms persist.
   Explanation

   Number of half-lives elapsed:
   $$\text{number of half-lives}=\frac{t}{T_{1/2}}=\frac{20\mathrm{days}}{5\mathrm{days}}=4$$

   Remaining atoms:
   $$N=N_0{\left(\frac{1}{2}\right)}^4=1600\times \frac{1}{16}=100\text{atoms}$$

   The decay is exponential: $N(t)=N_0{e}^{-\lambda t}$ where $\lambda =\ln 2/T_{1/2}$. After each half-life, exactly half the remaining nuclei decay. Starting from 1600: after 5d → 800 → 400 → 200 → 100 ✓

   Key takeaway

   Radioactive decay: $N = N_0 (1/2)^{t/T_{1/2}}$; count the number of half-life periods, then apply $(1/2)^n$.

8. Question 8 · Medium

   According to special relativity, a muon traveling at $v=0.99c$ relative to Earth has a lifetime measured in Earth's frame that is longer than its proper lifetime. This phenomenon is called:

   • A
     Length contraction

     Why not A: Length contraction refers to the shortening of spatial dimensions in the direction of motion, not the extension of time intervals.
   • B
     Time dilationCorrect
   • C
     The twin paradox

     Why not C: The twin paradox is a thought experiment involving two observers; it is a consequence of time dilation but not the phenomenon itself.
   • D
     Mass-energy equivalence

     Why not D: $E=m{c}^2$ relates rest mass to energy; it does not directly describe the stretching of time intervals.
   Explanation

   Time dilation: moving clocks tick slower as observed from a stationary frame. The observed lifetime of the muon:
   $$\Delta t=\gamma \Delta t_0=\frac{\Delta t_0}{\sqrt{1-v^2/c^2}}$$

   At $v=0.99c$: $\gamma =1/\sqrt{1-0.99^2}=1/\sqrt{0.0199}\approx 7.1$

   Earth observers measure the muon's lifetime as $\approx 7.1\times$ longer than the proper lifetime measured in the muon's rest frame. This is experimentally confirmed — atmospheric muons created 15 km up routinely reach Earth's surface despite having a proper lifetime of only $\sim 2.2\mu \mathrm{s}$.

   Key takeaway

   Time dilation: moving clocks run slow; $\Delta t = \gamma\Delta t_0$ where $\gamma > 1$ for all $v > 0$.

9. Question 9 · Hard

   In the Compton effect, an X-ray photon of wavelength ${\lambda }_0$ scatters off a free electron and the scattered photon has a longer wavelength ${\lambda }^{\prime }$. The Compton shift $\Delta \lambda ={\lambda }^{\prime }-{\lambda }_0$ depends on:

   • A
     The initial wavelength ${\lambda }_0$ of the photon only

     Why not A: The Compton shift does not depend on the initial wavelength; it depends only on the scattering angle.
   • B
     The scattering angle $\theta$ only (and fundamental constants)Correct
   • C
     Both the initial wavelength and the scattering angle

     Why not C: The Compton shift formula $\Delta \lambda =(h/m_{e}c)(1-\cos \theta )$ contains no ${\lambda }_0$ term.
   • D
     The mass of the photon and the electron mass

     Why not D: Photons are massless; the Compton shift involves the electron mass $m_e$ but not a photon mass.
   Explanation

   The Compton shift formula:
   $$\Delta \lambda ={\lambda }^{\prime }-{\lambda }_0=\frac{h}{m_{e}c}(1-\cos \theta )$$

   where:

   • $h/m_{e}c\approx 2.43\times 10^{-12}\mathrm{m}=2.43\mathrm{pm}$ is the Compton wavelength of the electron
   • $\theta$ is the angle between scattered photon and original direction

   Key observations:

   • $\Delta \lambda =0$ for forward scattering ($\theta =0$)
   • $\Delta \lambda =2h/(m_{e}c)\approx 4.86\mathrm{pm}$ for backscattering ($\theta =180°$)
   • The shift is independent of ${\lambda }_0$ — same shift for any initial wavelength

   Compton scattering provided direct evidence for the particle nature of light (photons carry momentum).

   Key takeaway

   Compton shift: $\Delta\lambda = (h/m_ec)(1-\cos\theta)$; depends only on scattering angle, not initial wavelength.

10. Question 10 · Hard

    A nuclear reaction fuses two deuterium nuclei to produce helium-3 and a neutron:
    $${}_1^2\mathrm{H}+{}_1^2\mathrm{H}\to {}_2^3\mathrm{He}+{}_0^1\mathrm{n}$$
    Given that the mass defect is $\Delta m=3.3\times 10^{-30}\mathrm{kg}$, how much energy is released? ($c=3\times 10^{8}m/s$)

    • A
      $E\approx 2.97\times 10^{-13}\mathrm{J}$Correct
    • B
      $E\approx 9.9\times 10^{-22}\mathrm{J}$

      Why not B: This uses $E=\Delta m\cdot c$ instead of $E=\Delta m\cdot c^2$; must square the speed of light.
    • C
      $E\approx 3.0\times 10^{-13}\mathrm{J}$

      Why not C: This rounds $\Delta m\cdot c^2$ excessively; the correct value is closer to $2.97\times 10^{-13}\mathrm{J}$.
    • D
      $E\approx 9.9\times 10^{-13}\mathrm{J}$

      Why not D: This may use $c=3\times 10^9$ (factor of 10 error) or miscalculate $c^2$.
    Explanation

    Mass-energy equivalence ($E=m{c}^2$):
    $$E=\Delta m\cdot c^2=(3.3\times 10^{-30}\mathrm{kg})(3\times 10^{8}m/s{)}^2$$
    $$E=3.3\times 10^{-30}\times 9\times 10^{16}=29.7\times 10^{-14}=2.97\times 10^{-13}\mathrm{J}$$

    Converting to MeV: $E=2.97\times 10^{-13}/(1.6\times 10^{-13})\approx 1.86\mathrm{MeV}$.

    This is a fusion reaction — the products are more tightly bound than the reactants, releasing energy. Fusion powers the sun and hydrogen bombs.

    Key takeaway

    Nuclear energy release: $E = \Delta m \cdot c^2$; compute $c^2 = 9 \times 10^{16}\,\mathrm{m^2/s^2}$ carefully.

11. Question 11 · Hard

    According to the Heisenberg uncertainty principle, the minimum uncertainty in the momentum of an electron confined to a region of length $\Delta x=1.0\times 10^{-10}\mathrm{m}$ (atomic scale) is approximately: ($h=6.63\times 10^{-34}\mathrm{J}\cdot \mathrm{s}$)

    • A
      $\Delta p\approx 6.63\times 10^{-24}\mathrm{kg}\cdot m/s$

      Why not A: This uses $h/\Delta x$ directly instead of $h/(4\pi \Delta x)$ or $\hslash /\Delta x$; the uncertainty relation involves $\hslash /2$, not $h$.
    • B
      $\Delta p\approx 5.3\times 10^{-25}\mathrm{kg}\cdot m/s$Correct
    • C
      $\Delta p\approx 1.0\times 10^{-10}\mathrm{kg}\cdot m/s$

      Why not C: This simply uses $\Delta x$ as $\Delta p$ without applying the correct formula or constants.
    • D
      $\Delta p\approx 6.63\times 10^{-44}\mathrm{kg}\cdot m/s$

      Why not D: This would result from multiplying $h\times \Delta x$ instead of dividing.
    Explanation

    The Heisenberg uncertainty principle: $\Delta x\cdot \Delta p\ge \hslash /2$ where $\hslash =h/(2\pi )$.

    Minimum momentum uncertainty (using $\Delta x\cdot \Delta p\ge \hslash /2$):
    $$\Delta p\approx \frac{\hslash }{2\Delta x}=\frac{h}{4\pi \Delta x}$$
    $$\Delta p\approx \frac{6.63\times 10^{-34}}{4\pi \times 1.0\times 10^{-10}}=\frac{6.63\times 10^{-34}}{1.257\times 10^{-9}}\approx 5.3\times 10^{-25}\mathrm{kg}\cdot m/s$$

    For an electron ($m_e=9.11\times 10^{-31}\mathrm{kg}$), this corresponds to speed $\Delta v=\Delta p/m_e\approx 5.8\times 10^{5}m/s$ — significant quantum uncertainty at the atomic scale.

    Key takeaway

    Heisenberg: $\Delta x \cdot \Delta p \geq \hbar/2 = h/(4\pi)$; confining a particle tightly forces large momentum uncertainty.

12. Question 12 · Hard

    A particle is accelerated through a potential difference of $V=50\mathrm{kV}$. Using the de Broglie relation $\lambda =h/p$ and energy conservation (non-relativistic), what is the de Broglie wavelength of a proton accelerated through this potential? ($m_p=1.67\times 10^{-27}\mathrm{kg}$, $e=1.6\times 10^{-19}\mathrm{C}$, $h=6.63\times 10^{-34}\mathrm{J}\cdot \mathrm{s}$)

    • A
      $\lambda \approx 1.28\times 10^{-13}\mathrm{m}$Correct
    • B
      $\lambda \approx 5.5\times 10^{-12}\mathrm{m}$

      Why not B: This may result from using the electron mass instead of the proton mass; electrons are $\sim 1836$ times lighter, giving a longer wavelength.
    • C
      $\lambda \approx 1.28\times 10^{-10}\mathrm{m}$

      Why not C: This is the proton wavelength off by $10^3$; possibly from using $V=50\mathrm{V}$ instead of $50\mathrm{kV}$.
    • D
      $\lambda \approx 6.63\times 10^{-34}\mathrm{m}$

      Why not D: This is Planck's constant itself, not the wavelength; the wavelength is $h/p$, not just $h$.
    Explanation

    Step 1: Find the kinetic energy gained by the proton:
    $$KE=eV=(1.6\times 10^{-19})(50\times 10^3)=8.0\times 10^{-15}\mathrm{J}$$

    Step 2: Find the momentum (non-relativistic: $KE=p^2/(2m)$):
    $$p=\sqrt{2m_p\cdot KE}=\sqrt{2(1.67\times 10^{-27})(8.0\times 10^{-15})}$$
    $$p=\sqrt{2.672\times 10^{-41}}=\sqrt{26.72\times 10^{-42}}\approx 5.17\times 10^{-21}\mathrm{kg}\cdot m/s$$

    Step 3: de Broglie wavelength:
    $$\lambda =\frac{h}{p}=\frac{6.63\times 10^{-34}}{5.17\times 10^{-21}}\approx 1.28\times 10^{-13}\mathrm{m}$$

    This is on the order of a nuclear diameter ($\sim 10^{-15}\mathrm{m}$), much smaller than an atom ($\sim 10^{-10}\mathrm{m}$). High-energy protons can probe nuclear structure.

    Key takeaway

    Accelerated particle: $KE = eV$, then $p = \sqrt{2mKE}$, then $\lambda = h/p$; three-step chain.