These are original Marlbridge practice questions, not past exam questions. The mark points are a suggested Marlbridge rubric to help you check your work; they are not an official scoring guideline. Data for every question: k = 9.0 × 10⁹ N·m²/C²; e = 1.60 × 10⁻¹⁹ C; h = 6.63 × 10⁻³⁴ J·s; electron mass mₑ = 9.11 × 10⁻³¹ kg; proton and neutron mass 1.67 × 10⁻²⁷ kg; hydrogen ground-state Bohr radius r₁ = 5.29 × 10⁻¹¹ m. Treat the nucleus as fixed. All speeds are much less than c. A scientific calculator is assumed. Round only at the end.
Question 1 (multiple choice · foundation)
How many protons, neutrons and electrons are there in the ion ³⁷₁₇Cl⁻?
- (A) 17 protons, 20 neutrons, 18 electrons
- (B) 17 protons, 20 neutrons, 16 electrons
- (C) 17 protons, 37 neutrons, 18 electrons
- (D) 20 protons, 17 neutrons, 21 electrons
Answer and explanation
Answer: (A). Z = 17, so there are 17 protons. N = A − Z = 37 − 17 = 20 neutrons. A 1− charge means one more electron than protons: 18 electrons.
- (B) treats the negative ion as having lost an electron. A negative ion has gained electrons.
- (C) uses the mass number as the number of neutrons. A counts protons and neutrons.
- (D) swaps the roles of the two numbers. The bottom-left number, Z, is the number of protons.
Question 2 (multiple choice · core)
In each pair, X and Y stand for chemical symbols, which may or may not be the same. Which pair of neutral atoms are isotopes of the same element?
- (A) ²⁴₁₂X and ²⁶₁₂Y
- (B) ⁴⁰₁₈X and ⁴⁰₂₀Y
- (C) ¹⁴₆X and ¹⁴₇Y
- (D) ²³₁₁X and ²⁴₁₂Y
Answer and explanation
Answer: (A). Both atoms have Z = 12, so they are the same element (X and Y must be the same symbol). Their mass numbers differ, so they have different numbers of neutrons (12 and 14). That is the definition of isotopes.
- (B) has the same mass number but different Z (18 and 20), so these are two different elements.
- (C) is the same kind of error: equal A, but Z = 6 and Z = 7 are different elements.
- (D) has the same number of neutrons (12 in each) but different Z, so again two different elements.
Question 3 (multiple choice · core)
In the Bohr model of hydrogen, the n = 2 orbit has 4 times the radius of the n = 1 orbit. What is the ratio (electric force on the electron in n = 2) ÷ (electric force on the electron in n = 1)?
- (A) 1/16
- (B) 1/4
- (C) 1/2
- (D) 4
Answer and explanation
Answer: (A). Coulomb’s law gives F ∝ 1/r². With r multiplied by 4, F is multiplied by 1/4² = 1/16.
- (B) uses F ∝ 1/r, which is how the electric potential energy depends on r, not the force.
- (C) is the ratio of the speeds (v ∝ 1/n), not of the forces.
- (D) makes the force grow with distance. The attraction gets weaker as the electron moves further from the nucleus.
Question 4 (multiple choice · core)
According to the standing-wave picture of the Bohr model, why can an electron in a hydrogen atom have only certain orbits?
- (A) The orbit’s circumference must be a whole number of the electron’s de Broglie wavelengths.
- (B) The orbit’s radius must be a whole number of the electron’s de Broglie wavelengths.
- (C) The electric force must be balanced by an equal outward force on the electron.
- (D) The electron must emit a photon on every orbit, and only some orbits give whole photons.
Answer and explanation
Answer: (A). The electron’s wave must join itself smoothly after one trip around the orbit. That only happens when 2πr = nλ for a whole number n.
- (B) uses the radius instead of the distance around the orbit. The wave travels along the circumference.
- (C) is a misunderstanding of circular motion: the net force is not zero; the electric force provides the centripetal force. In any case, this condition alone would allow any radius.
- (D) is wrong because an electron in an allowed orbit does not emit photons. Photons are emitted only when the atom changes state (Topic 15.3).
Question 5 (calculation · core)
A helium ion He⁺ has a nucleus of charge +2e and a single electron. In the Bohr model, its lowest orbit has radius 2.65 × 10⁻¹¹ m.
(a) Calculate the electric force on the electron. (b) Calculate the speed of the electron. (c) Calculate the electron’s de Broglie wavelength, and show that this orbit is the n = 1 orbit.
Worked solution
- (a) F = k(2e)(e)/r² = (9.0 × 10⁹)(2 × 1.60 × 10⁻¹⁹)(1.60 × 10⁻¹⁹) ÷ (2.65 × 10⁻¹¹)² = 6.56 × 10⁻⁷ N, towards the nucleus.
- (b) The electric force provides the centripetal force: F = mₑv²/r, so v = √(Fr/mₑ) = √(6.562 × 10⁻⁷ × 2.65 × 10⁻¹¹ ÷ 9.11 × 10⁻³¹) = 4.37 × 10⁶ m/s.
- (c) λ = h/(mₑv) = 6.63 × 10⁻³⁴ ÷ (9.11 × 10⁻³¹ × 4.369 × 10⁶) = 1.67 × 10⁻¹⁰ m.
- Circumference: 2πr = 2π × 2.65 × 10⁻¹¹ m = 1.67 × 10⁻¹⁰ m. So 2πr/λ = 1.00: exactly one wavelength fits, which is n = 1.
Suggested mark points (4): 1 for the force using charge 2e on the nucleus; 1 for setting F equal to mv²/r and getting v; 1 for λ; 1 for comparing 2πr with λ and concluding n = 1.
Common error: using charge e for the nucleus. That gives F = 3.28 × 10⁻⁷ N and v = 3.09 × 10⁶ m/s, and then the circumference does not hold a whole number of wavelengths, which is a useful warning sign.
Question 6 (constructed response · core)
A neutral lithium atom is written ⁷₃Li. It can lose one electron to become the ion ⁷₃Li⁺.
(a) State the numbers of protons, neutrons and electrons in the neutral atom and in the ion. (b) Calculate the net charge of the ion in coulombs. (c) Estimate the mass of the neutral atom. By what percentage does the mass change when it becomes the ion? Explain why the mass of an atom is dominated by its nucleus. (d) Another lithium atom is written ⁶₃Li. A student says: “⁶Li and ⁷Li are different elements because they have different mass numbers.” Evaluate this claim, and state which property of the atom mainly decides how it reacts with other atoms.
Worked solution and suggested Marlbridge rubric
(a) Atom: 3 protons, 7 − 3 = 4 neutrons, 3 electrons. Ion: 3 protons, 4 neutrons, 2 electrons.
(b) Net charge = (3 − 2)e = +1.60 × 10⁻¹⁹ C.
(c) Nucleus: 7 × 1.67 × 10⁻²⁷ kg = 1.169 × 10⁻²⁶ kg; adding 3 electrons gives about 1.17 × 10⁻²⁶ kg. Losing one electron removes 9.11 × 10⁻³¹ kg, which is 9.11 × 10⁻³¹ ÷ 1.169 × 10⁻²⁶ = 7.8 × 10⁻⁵, about 0.008%. The nucleus dominates because each proton and neutron is about 1800 times as massive as an electron, and the atom has more nucleons than electrons.
(d) The claim is wrong. Both atoms have Z = 3, and the number of protons decides the element, so both are lithium. They are different isotopes (3 and 4 neutrons). How an atom reacts depends mainly on the number and arrangement of its electrons, which are the same for the two neutral isotopes.
| Point | What earns it |
|---|---|
| 1 | Correct counts for the atom (3, 4, 3) |
| 1 | Correct counts for the ion, with 2 electrons, and charge +1.60 × 10⁻¹⁹ C |
| 1 | Atom mass ≈ 1.17 × 10⁻²⁶ kg and a percentage change of order 0.01% or less |
| 1 | Explains nuclear dominance using the mass ratio of nucleons to electrons |
| 1 | Rejects the claim because Z (protons) identifies the element; identifies them as isotopes |
| 1 | States that the electrons (number and arrangement) mainly decide how the atom interacts |
Accept 0.0078% or 7.8 × 10⁻⁵ as a fraction. Do not award the (d) claim point for “they are isotopes” with no reference to Z.
Question 7 (constructed response · stretch)
In the Bohr model of hydrogen, every allowed orbit must satisfy two conditions:
- the force condition, ke²/r² = mₑv²/r, and
- the standing-wave condition, 2πr = nλ with λ = h/(mₑv).
(a) Use the force condition to show that v ∝ 1/√r, and hence that λ ∝ √r. (b) Using your result from (a), show that the number of wavelengths that fit around an orbit, 2πr/λ, is proportional to √r. Hence explain why the allowed radii are r = n²r₁. (c) A student suggests an orbit of radius 2.5r₁. Determine how many de Broglie wavelengths would fit around it, and decide whether the orbit is allowed. (d) The total energy of an orbit is E = −ke²/(2r). Determine the ratio E₂/E₁ for the n = 2 and n = 1 orbits, and state which orbit is more tightly bound.
Worked solution and suggested Marlbridge rubric
(a) From ke²/r² = mₑv²/r, v² = ke²/(mₑr), so v = √(ke²/(mₑr)) ∝ 1/√r. Then λ = h/(mₑv) ∝ 1/v ∝ √r.
(b) 2πr/λ ∝ r/√r = √r. For the n = 1 orbit, exactly one wavelength fits, so the number that fits at radius r is √(r/r₁). Setting this equal to a whole number n gives √(r/r₁) = n, so r = n²r₁.
(c) Number of wavelengths = √(2.5) = 1.58. This is not a whole number, so the wave would not join up with itself. The orbit is not allowed. (Check of the method: for r = 4r₁, √4 = 2 wavelengths, which is the allowed n = 2 orbit.)
(d) E ∝ 1/r, and r₂ = 4r₁, so E₂/E₁ = 1/4. Both are negative: E₁ = −13.6 eV and E₂ = −3.40 eV. The n = 1 orbit has the more negative energy, so it is more tightly bound: more energy is needed to free the electron from it.
| Point | What earns it |
|---|---|
| 1 | Derives v ∝ 1/√r from the force condition |
| 1 | Links λ ∝ 1/v to get λ ∝ √r |
| 1 | Shows 2πr/λ ∝ √r and uses n = 1 at r₁ to reach r = n²r₁ |
| 1 | (c) Finds about 1.58 wavelengths |
| 1 | (c) Concludes “not allowed” because the number is not a whole number |
| 1 | (d) E₂/E₁ = 1/4 and identifies n = 1 as more tightly bound, with reason |
Accept a full algebraic route in (b) that eliminates v between the two conditions to get r = n²h²/(4π²kmₑe²). Accept, in (c), a numerical route: v = v₁/√2.5, λ = λ₁√2.5, and 2π(2.5r₁)/(λ₁√2.5) = √2.5.
How did you do?
- Q1 or Q6 wrong: re-read “Elements, isotopes and ions” and Worked example 1 in the study guide.
- Q2 wrong: remember that only Z decides the element.
- Q3 or Q5 wrong: revisit “The Bohr model” and Worked example 2; write F = k|q₁q₂|/r² = mv²/r first.
- Q4 wrong: look again at Figure 2 and the standing-wave condition.
- Q7 incomplete: work through Worked example 3, then redo the proportional reasoning.
Then tick off the topic checklist.