Modern Physics: Unit Diagnostic (Physics 2 Unit 15)

What this diagnostic is for

Use this check to find which topics in Unit 15 (Modern Physics) to revisit. It has eight multiple-choice and three short-answer questions. These are original Marlbridge practice questions, not past exam questions. They are not calibrated against real exam results, and your total does not predict a score. What matters is which questions you miss.

How to take it:

Work without notes for about 30 minutes, and write a reason for each answer.
A scientific calculator is allowed. Data: hc = 1240 eV·nm; 1 eV = 1.60 × 10⁻¹⁹ J; mₑ = 9.11 × 10⁻³¹ kg; neutron mass 1.67 × 10⁻²⁷ kg; h/(mₑc) = 2.43 pm.
Open each answer only after you have committed to your own.

Question 1 (multiple choice · 15.1)

A neutron and an electron have the same de Broglie wavelength. Which statement is correct?

(A) They have the same speed, and the neutron has the larger kinetic energy.
(B) They have the same momentum and the same kinetic energy.
(C) They have the same momentum, and the electron has the larger kinetic energy.
(D) The neutron has the larger momentum, because it has the larger mass.

Answer and explanation

Answer: (C). λ = h/p, so the same λ means the same momentum. With equal p, K = p²/(2m), so the electron has about 1833 times more kinetic energy.

(A) treats λ as depending on speed alone. It depends on mv.
(B) forgets that K depends on mass too.
(D) uses mass instead of momentum. A heavier particle with the same λ moves more slowly.

If you missed this: read “Matter waves” in the Topic 15.1 study guide.

Question 2 (multiple choice · 15.2)

In the Bohr model of hydrogen, which row compares the n = 4 orbit correctly with the n = 2 orbit?

(A) Radius × 2, speed × ½, de Broglie wavelength × 2
(B) Radius × 4, speed × ½, de Broglie wavelength × 2
(C) Radius × 4, speed × 2, de Broglie wavelength × ½
(D) Radius × 4, speed × ½, de Broglie wavelength unchanged

Answer and explanation

Answer: (B). Allowed radii are r = n²r₁, so r is multiplied by (4/2)² = 4. The force condition gives v ∝ 1/√r, so v halves. Then λ = h/(mₑv) doubles.

(A) uses r ∝ n instead of n².
(C) makes the electron faster where the force is weaker.
(D) forgets that λ = h/(mₑv) changes when v changes.

If you missed this: read “Why only some orbits: the standing-wave picture” in the Topic 15.2 study guide.

Question 3 (multiple choice · 15.3)

Atom Q (also used in Question 9) is a single-electron atom with levels n = 1: −11.0 eV; n = 2: −4.8 eV; n = 3: −2.4 eV; n = 4: −1.2 eV. Light with every wavelength from 100 nm to 1100 nm passes through a cool gas of atom Q, all in the ground state. At which wavelength is there a dark line?

(A) 1033 nm
(B) 517 nm
(C) 344 nm
(D) 200 nm

Answer and explanation

Answer: (D). A ground-state atom absorbs only photons that take it from n = 1 to another level. The 1 → 2 gap is 6.2 eV, and 1240 ÷ 6.2 = 200 nm.

(A) is the 3 → 4 gap (1.2 eV), which does not start from the ground state.
(B) is the 2 → 3 gap (2.4 eV): an emission line, but a cool gas has almost no atoms in n = 2.
(C) is the 2 → 4 gap (3.6 eV), also from an excited state.

If you missed this: read “Line spectra and what they tell you” in the Topic 15.3 study guide.

Question 4 (multiple choice · 15.4)

A blackbody is heated. Its peak wavelength moves from 1000 nm to 800 nm. Its surface area does not change. By what factor does the total power it emits increase?

(A) 2.44
(B) 1.25
(C) 0.41
(D) 1.56

Answer and explanation

Answer: (A). Wien’s law: λ_max ∝ 1/T, so T rises by 1000/800 = 1.25. The Stefan–Boltzmann law: P ∝ AT⁴, so P rises by 1.25⁴ = 2.44.

(B) is the temperature factor, as if P ∝ T.
(C) inverts the ratio. A shorter peak means a hotter body and more power.
(D) uses T² instead of T⁴.

If you missed this: read “Wien’s law: the peak wavelength” and “The Stefan–Boltzmann law: total power” in the Topic 15.4 study guide.

Question 5 (multiple choice · 15.5)

Light of frequency f ejects electrons from a metal with maximum kinetic energy 1.0 eV. Light of frequency 2f gives a maximum kinetic energy of 4.0 eV with the same metal. What is the work function of the metal?

(A) 1.0 eV
(B) 3.0 eV
(C) 2.0 eV
(D) 4.0 eV

Answer and explanation

Answer: (C). K_max = hf − φ. So hf − φ = 1.0 eV and 2hf − φ = 4.0 eV. Subtracting gives hf = 3.0 eV, so φ = 3.0 − 1.0 = 2.0 eV.

(A) is K_max at the lower frequency, the energy left over.
(B) is the photon energy hf at the lower frequency, not the work function.
(D) is K_max at the higher frequency.

If you missed this: read “The photon explanation” in the Topic 15.5 study guide.

Question 6 (multiple choice · 15.6)

An X-ray photon scatters off a free electron and is detected at 90° to its original direction. Its wavelength is now 3.0% longer than before. What was its wavelength before the collision?

(A) 162 pm
(B) 81.0 pm
(C) 0.0729 pm
(D) 117 pm

Answer and explanation

Answer: (B). At 90°, Δλ = (2.43 pm)(1 − cos 90°) = 2.43 pm. This is 3.0% of the starting wavelength, so λ = 2.43 pm ÷ 0.030 = 81.0 pm.

(A) uses the 180° shift, 4.86 pm.
(C) multiplies by 0.030 instead of dividing.
(D) comes from a calculator in radian mode, which gives cos 90 = −0.448.

If you missed this: read “The Compton shift equation” in the Topic 15.6 study guide.

Question 7 (multiple choice · 15.7)

Two light nuclei fuse into one heavier nucleus and a neutron, and energy is released. Which statement is correct?

(A) Mass is conserved; the released energy comes from the kinetic energy the nuclei had before.
(B) The products have more total mass than the reactants, because energy was released.
(C) Some nucleons disappear, and their mass becomes the released energy.
(D) The products have less total mass than the reactants, and they are more tightly bound.

Answer and explanation

Answer: (D). Total energy, including rest energy, is conserved. The products gain kinetic energy, so their rest energy and mass are lower (E = Δmc²): they are more tightly bound.

(A) Mass is not conserved in nuclear reactions. Nucleon number is.
(B) is backwards: released energy comes from a decrease in mass.
(C) breaks conservation of nucleon number. No nucleons disappear.

If you missed this: read “Mass and energy” in the Topic 15.7 study guide.

Question 8 (multiple choice · 15.8)

A nucleus with Z = 84 and A = 200 undergoes beta-plus decay. The daughter then undergoes alpha decay. What are Z and A of the final nucleus?

(A) Z = 81, A = 196
(B) Z = 83, A = 196
(C) Z = 81, A = 195
(D) Z = 82, A = 196

Answer and explanation

Answer: (A). Beta-plus: a proton becomes a neutron, so Z falls by 1 and A stays 200 (Z = 83). Alpha: Z falls by 2 and A by 4. Final: Z = 81, A = 196.

(B) raises Z, as beta-minus would.
(C) takes one nucleon away for the positron. A positron is not a nucleon.
(D) forgets the beta-plus decay.

If you missed this: read “The four types of decay” in the Topic 15.8 study guide.

Question 9 (short answer · 15.3)

Use atom Q from Question 3.

(a) A ground-state atom absorbs a photon of wavelength 144 nm. Which level does it reach? (b) It returns to the ground state in two steps. Calculate the wavelength of each photon it emits. (c) Calculate the minimum photon energy that can ionize an atom in the n = 2 state. Explain why this is less than for a ground-state atom.

Worked answer

(a) E = 1240 ÷ 144 = 8.61 eV. Then −11.0 + 8.6 = −2.4 eV, so the atom reaches n = 3.
(b) 3 → 2: 2.4 eV, so λ = 1240 ÷ 2.4 = 517 nm. 2 → 1: 6.2 eV, so λ = 200 nm. The two energies add to 8.6 eV, the energy absorbed.
(c) The electron must reach 0 eV: 0 − (−4.8 eV) = 4.8 eV. An atom in n = 2 is less tightly bound; from the ground state 11.0 eV is needed.

If you missed this: read “Energy level diagrams” and “Binding energy and ionization” in the Topic 15.3 study guide.

Question 10 (short answer · 15.5)

Light of wavelength 310 nm falls on a metal plate. The stopping potential is 1.50 V.

(a) Calculate the work function of the metal. (b) Calculate the threshold wavelength. (c) The intensity is tripled at the same wavelength. State what happens to the stopping potential and to the current before any stopping potential is applied.

Worked answer

(a) E = 1240 ÷ 310 = 4.00 eV. K_max = eV_s = 1.50 eV. So φ = 4.00 − 1.50 = 2.50 eV.
(b) λ₀ = 1240 ÷ 2.50 = 496 nm.
(c) The stopping potential is unchanged: each photon still has 4.00 eV. The current triples: three times as many photons per second free three times as many electrons.

If you missed this: read “Measuring K_max: the stopping potential” in the Topic 15.5 study guide.

Question 11 (short answer · 15.7)

Invented isotopes X and Y have half-lives of 5.0 min and 10.0 min. At t = 0 a sample holds 1.2 × 10⁸ undecayed nuclei of X and 4.0 × 10⁷ of Y.

(a) Calculate the decay constant of X, in min⁻¹ and in s⁻¹. (b) Calculate the time at which the numbers of undecayed X and Y nuclei are equal. (c) Calculate how many nuclei of each remain then.

Worked answer

(a) λ_X = ln 2 ÷ t½ = 0.693 ÷ 5.0 min = 0.139 min⁻¹ = 0.1386 ÷ 60 s = 2.31 × 10⁻³ s⁻¹.
(b) λ_Y = 0.0693 min⁻¹. Set 1.2 × 10⁸ e^(−λ_X t) = 4.0 × 10⁷ e^(−λ_Y t), so e^((λ_X − λ_Y)t) = 3 and t = ln 3 ÷ 0.0693 min⁻¹ = 15.8 min.
(c) e^(−λ_Y t) = 1/3, so N = 4.0 × 10⁷ ÷ 3 = 1.3 × 10⁷ of each. Check: X has (1/3)² = 1/9 left, and 1.2 × 10⁸ ÷ 9 = 1.3 × 10⁷.

If you missed this: read “Radioactive decay is random” and Worked example 2 in the Topic 15.7 study guide.

Your next step

Count a short answer as wrong if any part is missing, and any answer as wrong if it was a guess.

Topic	Question(s)	If you missed it, read
15.1 Quantum theory and wave-particle duality	1	Topic 15.1 study guide
15.2 The Bohr model of atomic structure	2	Topic 15.2 study guide
15.3 Emission and absorption spectra	3, 9	Topic 15.3 study guide
15.4 Blackbody radiation	4	Topic 15.4 study guide
15.5 The photoelectric effect	5, 10	Topic 15.5 study guide
15.6 Compton scattering	6	Topic 15.6 study guide
15.7 Fission, fusion and nuclear decay	7, 11	Topic 15.7 study guide
15.8 Types of radioactive decay	8	Topic 15.8 study guide

How to use your result

Start with the topic where you missed the most. Read its study guide, then do that topic’s practice questions.
Look at why you missed it. Each explanation names the mistake behind a distractor. A unit slip (eV and joules, radian mode) needs a habit, not a re-read.
Missed nothing? Go to the Unit 15 mixed review, where questions combine topics.
Come back to this check after a week of revision and answer again before opening the explanations.