FAD1022 CQ7: Rapid-Fire Drill Pack — Atomic Physics

Objective: Master all Atomic Physics concepts for CQ7.
Target: 3-4 min per problem (CQ7 style: 4 questions in 15 min).
Total problems: 28
Estimated time: ~100 min

Cheat Sheet (Memorize First)

Constants for CQ7

Constant Symbol Value
Planck's constant $h$ $6.63 \times 10^{-34}\ \text{J·s}$
Reduced Planck's constant $\hbar$ $1.055 \times 10^{-34}\ \text{J·s}$
Speed of light $c$ $3.00 \times 10^{8}\ \text{m/s}$
Elementary charge $e$ $1.60 \times 10^{-19}\ \text{C}$
Electron mass $m_e$ $9.11 \times 10^{-31}\ \text{kg}$
Rydberg constant $R_H$ $1.097 \times 10^{7}\ \text{m}^{-1}$
Useful: $hc$ $hc$ $1240\ \text{eV·nm}$

Bohr Model

Quantity Formula
Radius of orbit $n$ $r_n = n^2 a_0$
Bohr radius $a_0 = 0.529\ \text{Å} = 5.29 \times 10^{-11}\ \text{m}$
Energy level $n$ $E_n = -\frac{13.6}{n^2}\ \text{eV}$
Angular momentum $L_n = n\hbar$
Velocity in orbit $n$ $v_n = \frac{v_1}{n}$ where $v_1 = \alpha c \approx c/137$

Photon Energy & Spectra

Quantity Formula
Photon energy $E = hf = \frac{hc}{\lambda}$
Transition energy $\Delta E = E_i - E_f = hf = \frac{hc}{\lambda}$
Rydberg formula $\frac{1}{\lambda} = R_H(\frac{1}{n_f^2} - \frac{1}{n_i^2})$

Hydrogen Spectral Series

Series Transitions Region
Lyman $n \ge 2 \to n = 1$ UV
Balmer $n \ge 3 \to n = 2$ Visible
Paschen $n \ge 4 \to n = 3$ IR
Brackett $n \ge 5 \to n = 4$ IR

Photoelectric Effect

Quantity Formula
Einstein's equation $K_{\text{max}} = hf - \phi$
Stopping potential $eV_s = K_{\text{max}}$
Threshold frequency $f_0 = \frac{\phi}{h}$
Cutoff wavelength $\lambda_c = \frac{hc}{\phi}$
Work function $\phi = hf_0$

de Broglie Wavelength

Quantity Formula
Wavelength $\lambda = \frac{h}{p} = \frac{h}{mv}$
For electrons $\lambda = \frac{h}{\sqrt{2m_e eV}} = \frac{1.227}{\sqrt{V}}\ \text{nm}$ (V in volts)

Part A: Bohr Model — Energy Levels & Transitions (8 problems)

Target: 3 min per problem.

Set A1 — Energy Level Calculations (4 problems)

  1. Calculate the energy of the first four energy levels ($n = 1, 2, 3, 4$) for hydrogen in eV.

  2. Calculate the ionization energy of hydrogen from: (a) The ground state ($n = 1$) (b) The first excited state ($n = 2$)

  3. A hydrogen atom is excited to the $n = 4$ level. Calculate: (a) The energy of the atom in this state (b) The minimum energy required to ionize the atom from this state (c) How many different spectral lines can be emitted as the atom returns to the ground state?

  4. Calculate the binding energy of an electron in the $n = 3$ state of hydrogen. Express your answer in both eV and Joules.

Score: ___/4

Set A2 — Spectral Transitions (4 problems)

  1. Calculate the wavelength and frequency of the photon emitted when a hydrogen atom transitions from $n = 3$ to $n = 2$ (first Balmer line, H-alpha).

  2. A hydrogen atom emits a photon with wavelength $\lambda = 486\ \text{nm}$. Identify the initial and final energy levels for this transition (H-beta line).

  3. Calculate the shortest wavelength photon that can be emitted in the: (a) Lyman series (b) Balmer series (c) Paschen series

  4. What is the minimum energy photon that can ionize a hydrogen atom in the $n = 2$ state? What is its wavelength?

Score: ___/4

Part B: Bohr Model — Radii & Velocities (6 problems)

Target: 3 min per problem.

Set B1 — Orbital Radii & Properties (3 problems)

  1. Calculate the radius of the first three Bohr orbits for hydrogen.

  2. The radius of a Bohr orbit in hydrogen is measured to be $r = 4.76\ \text{Å}$. Identify the quantum number $n$ for this orbit.

  3. Calculate the ratio of the electron's speed in the $n = 2$ orbit to its speed in the $n = 1$ orbit for hydrogen.

Score: ___/3

Set B2 — Angular Momentum & Velocity (3 problems)

  1. Calculate the angular momentum of an electron in the $n = 3$ state of hydrogen.

  2. The Bohr quantization condition states $L = n\hbar$. Verify this by calculating the angular momentum using $L = m_e v_n r_n$ for $n = 1$.

  3. Calculate the speed of an electron in the $n = 1$ state of hydrogen. What fraction of the speed of light is this?

Score: ___/3

Part C: Photoelectric Effect (8 problems)

Target: 3-4 min per problem.

Set C1 — Basic Photoelectric Calculations (4 problems)

  1. A metal has a work function $\phi = 2.50\ \text{eV}$. Calculate: (a) The threshold frequency (b) The cutoff wavelength (c) The maximum kinetic energy of emitted electrons when light with $\lambda = 400\ \text{nm}$ is incident

  2. Light of wavelength $\lambda = 300\ \text{nm}$ is incident on a metal surface with work function $\phi = 3.20\ \text{eV}$. Calculate: (a) The photon energy in eV (b) The maximum kinetic energy of photoelectrons (c) The stopping potential

  3. Photoelectrons are emitted from a metal surface with a maximum kinetic energy of $K_{\text{max}} = 1.80\ \text{eV}$ when illuminated by light with $\lambda = 350\ \text{nm}$. Calculate: (a) The work function of the metal (b) The threshold wavelength (c) Will light with $\lambda = 500\ \text{nm}$ cause photoemission?

  4. The stopping potential for photoelectrons from a metal is $V_s = 2.30\ \text{V}$ when illuminated by light with $\lambda = 250\ \text{nm}$. Calculate: (a) The maximum kinetic energy (b) The work function (c) The threshold frequency

Score: ___/4

Set C2 — Photoelectric Graphs & Interpretation (4 problems)

  1. Sketch a graph of $K_{\text{max}}$ versus frequency $f$ for the photoelectric effect. Label the slope, y-intercept, and x-intercept. What physical quantities do these represent?

  2. For a given metal, explain what happens to: (a) $K_{\text{max}}$ if the light intensity is doubled while frequency remains constant (b) The photocurrent if the light intensity is doubled (c) $K_{\text{max}}$ if the frequency is increased

  3. Two metals A and B have work functions $\phi_A = 2.0\ \text{eV}$ and $\phi_B = 3.5\ \text{eV}$. On the same graph, sketch $K_{\text{max}}$ versus $f$ for both metals. Label which is which.

  4. Explain why there is a threshold frequency in the photoelectric effect but not in the classical wave theory of light.

Score: ___/4

Part D: de Broglie Wavelength & Wave-Particle Duality (6 problems)

Target: 3 min per problem.

  1. Calculate the de Broglie wavelength of: (a) An electron moving at $v = 1.0 \times 10^6\ \text{m/s}$ (b) A proton moving at $v = 1.0 \times 10^6\ \text{m/s}$ (c) A 60 kg person running at $v = 5.0\ \text{m/s}$

  2. An electron is accelerated from rest through a potential difference of $V = 100\ \text{V}$. Calculate its de Broglie wavelength.

  3. Calculate the accelerating voltage needed to give an electron a de Broglie wavelength of $\lambda = 0.10\ \text{nm}$.

  4. In an electron microscope, electrons are accelerated through $V = 50\ \text{kV}$. Calculate: (a) The de Broglie wavelength (b) The electron speed (c) Compare this wavelength to typical X-ray wavelengths

Score: ___/4

Final Scorecard

Part Topic Problems Raw Score
A — Bohr Energy Levels Sets A1-A2 8 ___/8
B — Bohr Radii & Velocity Sets B1-B2 6 ___/6
C — Photoelectric Effect Sets C1-C2 8 ___/8
D — de Broglie & Duality Set D 6 ___/6
TOTAL 28 ___/28

Proficiency Benchmarks for CQ7

Level Score Meaning
Exam-Ready $\ge 25/28$ (89%+) Ready for CQ7. Focus on speed.
Solid $22-24/28$ (79-86%) Good grasp. Review missed concepts.
Developing $17-21/28$ (61-75%) Drill weak areas again.
Needs Work $< 17/28$ (<61%) Re-study cheat sheet, retry tutorial.

Speed Benchmarks

  • < 80 min: Excellent speed for CQ7 conditions.
  • 80–100 min: Good. Practice under time pressure.
  • > 100 min: Need more pattern recognition drills.

Error Log Template

After grading, list every wrong problem with a one-word reason:

Problem Reason

Re-solve all wrong problems immediately with notes, then again in 24 hours without notes.

Answer Key

Part A — Bohr Energy Levels

  1. $E_n = -13.6/n^2$

    • $n = 1$: $E_1 = -13.6\ \text{eV}$
    • $n = 2$: $E_2 = -3.40\ \text{eV}$
    • $n = 3$: $E_3 = -1.51\ \text{eV}$
    • $n = 4$: $E_4 = -0.850\ \text{eV}$

  2. (a) $\Delta E = 0 - (-13.6) = 13.6\ \text{eV}$ (b) $\Delta E = 0 - (-3.40) = 3.40\ \text{eV}$

  3. (a) $E_4 = -0.850\ \text{eV}$ (b) $0.850\ \text{eV}$ (c) Number of transitions = $\frac{n(n-1)}{2} = \frac{4(3)}{2} = 6$

  4. Binding energy = $|E_3| = 1.51\ \text{eV} = 1.51 \times 1.6 \times 10^{-19} = 2.42 \times 10^{-19}\ \text{J}$

  5. $\Delta E = E_3 - E_2 = -1.51 - (-3.40) = 1.89\ \text{eV}$ $\lambda = \frac{hc}{\Delta E} = \frac{1240}{1.89} = 656\ \text{nm}$ (red) $$f = \frac{c}{\lambda} = \frac{3 \times 10^8}{656 \times 10^{-9}} = 4.57 \times 10^{14}\ \text{Hz}$$

  6. $E = \frac{hc}{\lambda} = \frac{1240}{486} = 2.55\ \text{eV}$ Transition: $n = 4 \to n = 2$ ($\Delta E = -0.85 - (-3.40) = 2.55\ \text{eV}$)

  7. (a) Lyman limit: $n = \infty \to n = 1$: $\lambda = \frac{hc}{13.6} = 91.2\ \text{nm}$ (b) Balmer limit: $n = \infty \to n = 2$: $\lambda = \frac{hc}{3.40} = 365\ \text{nm}$ (c) Paschen limit: $n = \infty \to n = 3$: $\lambda = \frac{hc}{1.51} = 821\ \text{nm}$

  8. Minimum energy = $|E_2| = 3.40\ \text{eV}$ $$\lambda = \frac{hc}{3.40} = 365\ \text{nm}$$

Part B — Bohr Radii

  1. $r_n = n^2 a_0 = n^2(0.529\ \text{Å})$

    • $n = 1$: $r_1 = 0.529\ \text{Å} = 5.29 \times 10^{-11}\ \text{m}$
    • $n = 2$: $r_2 = 2.12\ \text{Å} = 2.12 \times 10^{-10}\ \text{m}$
    • $n = 3$: $r_3 = 4.76\ \text{Å} = 4.76 \times 10^{-10}\ \text{m}$

  2. $r = n^2 a_0$ → $n = \sqrt{r/a_0} = \sqrt{4.76/0.529} = \sqrt{9} = 3$

  3. $v_n = v_1/n$, so $\frac{v_2}{v_1} = \frac{1}{2}$

  4. $L = n\hbar = 3(1.055 \times 10^{-34}) = 3.17 \times 10^{-34}\ \text{J·s}$

  5. $v_1 = \frac{e^2}{2\varepsilon_0 h} = 2.19 \times 10^6\ \text{m/s}$, $r_1 = 5.29 \times 10^{-11}\ \text{m}$ $L = m_e v_1 r_1 = (9.11 \times 10^{-31})(2.19 \times 10^6)(5.29 \times 10^{-11}) = 1.055 \times 10^{-34} = \hbar$ ✓

  6. $v_1 = 2.19 \times 10^6\ \text{m/s}$ $$\frac{v_1}{c} = \frac{2.19 \times 10^6}{3 \times 10^8} = 7.3 \times 10^{-3} \approx 1/137$$

Part C — Photoelectric Effect

  1. (a) $f_0 = \phi/h = (2.50 \times 1.6 \times 10^{-19})/(6.63 \times 10^{-34}) = 6.03 \times 10^{14}\ \text{Hz}$ (b) $\lambda_c = hc/\phi = 1240/2.50 = 496\ \text{nm}$ (c) $E = hc/\lambda = 1240/400 = 3.10\ \text{eV}$ $$K_{\text{max}} = 3.10 - 2.50 = 0.60\ \text{eV}$$

  2. (a) $E = 1240/300 = 4.13\ \text{eV}$ (b) $K_{\text{max}} = 4.13 - 3.20 = 0.93\ \text{eV}$ (c) $V_s = K_{\text{max}}/e = 0.93\ \text{V}$

  3. (a) $E = 1240/350 = 3.54\ \text{eV}$ $$\phi = E - K_{\text{max}} = 3.54 - 1.80 = 1.74\ \text{eV}$$ (b) $\lambda_c = 1240/1.74 = 713\ \text{nm}$ (c) $E(500\ \text{nm}) = 1240/500 = 2.48\ \text{eV} > 1.74\ \text{eV}$, so yes

  4. (a) $K_{\text{max}} = eV_s = 2.30\ \text{eV}$ (b) $E = 1240/250 = 4.96\ \text{eV}$ $$\phi = 4.96 - 2.30 = 2.66\ \text{eV}$$ (c) $f_0 = \phi/h = (2.66 \times 1.6 \times 10^{-19})/(6.63 \times 10^{-34}) = 6.42 \times 10^{14}\ \text{Hz}$

  5. Graph: Straight line with positive slope = $h$, x-intercept = $f_0 = \phi/h$, y-intercept = $-\phi$

  6. (a) $K_{\text{max}}$ unchanged (depends only on frequency) (b) Photocurrent doubles (more photons → more electrons) (c) $K_{\text{max}}$ increases (linear relationship with frequency)

  7. Both have same slope $h$. Metal A has smaller x-intercept (smaller $\phi$), less negative y-intercept.

  8. In classical wave theory, energy is proportional to intensity, so any frequency should work given enough intensity. The photon model (quantum) explains threshold: each photon needs energy $hf \ge \phi$ to eject an electron.

Part D — de Broglie Wavelength

  1. (a) $\lambda = h/(m_e v) = (6.63 \times 10^{-34})/((9.11 \times 10^{-31})(10^6)) = 7.28 \times 10^{-10}\ \text{m} = 0.728\ \text{nm}$ (b) $\lambda = h/(m_p v) = (6.63 \times 10^{-34})/((1.67 \times 10^{-27})(10^6)) = 3.97 \times 10^{-13}\ \text{m}$ (c) $\lambda = h/(mv) = (6.63 \times 10^{-34})/(60 \times 5) = 2.21 \times 10^{-36}\ \text{m}$ (undetectably small)

  2. $\lambda = \frac{1.227}{\sqrt{V}}\ \text{nm} = \frac{1.227}{\sqrt{100}} = 0.123\ \text{nm} = 1.23\ \text{Å}$

  3. $\lambda = \frac{1.227}{\sqrt{V}}$ → $V = (1.227/\lambda)^2 = (1.227/0.10)^2 = 150\ \text{V}$

  4. (a) $\lambda = \frac{1.227}{\sqrt{50000}} = 5.48 \times 10^{-3}\ \text{nm} = 5.48\ \text{pm}$ (b) $eV = \frac{1}{2}m_e v^2$ → $v = \sqrt{2eV/m_e} = \sqrt{2(1.6 \times 10^{-19})(50000)/(9.11 \times 10^{-31})} = 1.33 \times 10^8\ \text{m/s}$ (c) X-ray wavelengths are typically 0.01 to 10 nm, so this is in the X-ray range

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