1True or false
An atom raised to an excited state normally stays there for many hours before returning to the ground state.
Show answer
Answer: False
Excited states usually last only about 10⁻⁸ s; the atom quickly drops back, emitting one or more photons.
!Common mistakeLearners often think excitation is permanent; that is why a gas glows only while energy is being supplied.
2True or false · ★ Challenge
In Bohr's model, an electron in a higher orbit moves faster than an electron in the ground state.
Show answer
Answer: False
In Bohr's model v ∝ 1/n: the electron moves fastest in the n = 1 orbit (about 2.2 × 10⁶ m/s) and more slowly in larger orbits, where the attraction of the nucleus is weaker.
!Common mistakeLearners often link 'higher energy' with 'faster'; the total energy rises with n, but the kinetic energy and speed fall.
3True or false · ★ Challenge
Thermionic emission and the photoelectric effect both release electrons from a metal surface, but they differ in where the electrons get the energy to escape.
Show answer
Answer: True
In thermionic emission the energy comes from heating the metal; in the photoelectric effect it comes from absorbed photons of light. In both cases the electron must gain at least the work function.
!Common mistakeLearners often think heating and shining light are the same process; they are two different ways of giving electrons the work function energy.
4True or false
Thomson's model was built after he showed that cathode rays are negatively charged particles much lighter than any atom.
Show answer
Answer: True
Thomson's discovery of the electron (1897) proved atoms have smaller parts, which led him to propose electrons embedded in a sphere of positive charge.
!Common mistakeLearners sometimes think the plum pudding model came from the gold-foil experiment; that experiment came later and disproved it.
5True or false · ★ Challenge
A proton and an electron accelerated from rest through the same potential difference gain the same kinetic energy, but the proton ends up moving much more slowly.
Show answer
Answer: True
Both carry charge of size e, so both gain eV of energy; with the same ½mv², the proton's much larger mass gives a much smaller speed.
!Common mistakeLearners often think the heavier particle gains more energy; the energy gained depends only on charge × p.d.
6True or false
An alpha particle heading straight towards a gold nucleus is turned back by electrostatic repulsion, before it ever touches the nucleus.
Show answer
Answer: True
Both are positive, so the repulsive force grows as the alpha particle approaches; its kinetic energy is changed into electric potential energy until it stops and is pushed back.
!Common mistakeLearners often imagine a mechanical collision like a ball hitting a wall; the turning back is caused by the electric force at a distance.
7True or false · ★ Challenge
In each hydrogen series the lines crowd closer and closer together towards the short-wavelength end, approaching a series limit.
Show answer
Answer: True
As n₂ increases, the upper levels get closer together, so successive lines differ less and less in wavelength, approaching the limit set by n₂ = ∞.
!Common mistakeLearners often expect equally spaced lines; the spacing shrinks because the energy levels themselves crowd together.
8Multiple choice · ★ Challenge
In Bohr's model, an electron in an allowed orbit
- Amoves in a straight line, so it does not accelerate
- Bis at rest, held by the attraction of the nucleus
- Cis accelerating towards the nucleus, yet gives out no radiation
- Dslowly loses energy by radiating and spirals towards the nucleus
Show answer
Answer: C. is accelerating towards the nucleus, yet gives out no radiation
Circular motion always involves centripetal acceleration; Bohr simply postulated that in the allowed orbits this acceleration does not produce radiation.
!Common mistakeChoosing 'spirals towards the nucleus' is the classical prediction that Bohr's postulate was designed to remove.
9True or false
A photon of blue light carries more energy than a photon of red light.
Show answer
Answer: True
E = hf = hc/λ; blue light has a shorter wavelength (higher frequency) than red light, so each blue photon has more energy.
!Common mistakeLearners often link 'red' with 'hot' and so with more energy; photon energy depends on frequency, not on colour associations.
10Multiple choice · ★ Challenge
If Thomson's plum pudding model had been correct, what would Geiger and Marsden have observed?
- AMost alpha particles would bounce straight back from the gold foil towards the source
- BAll alpha particles would be absorbed by the foil
- CThe alpha particles would be deflected into a few sharp beams
- DAlmost all would pass through with small deflections; none would bounce back
Show answer
Answer: D. Almost all would pass through with small deflections; none would bounce back
In Thomson's model the positive charge is spread thinly through the atom, so its electric field is weak everywhere; fast, heavy alpha particles would be deflected only slightly.
!Common mistakeChoosing 'bounce straight back' describes the surprising result that DISPROVED Thomson's model.
11Fill in the blank
The tiny central part of the atom that holds nearly all its mass and all its positive charge is called the ______.
Show answer
Answer: nucleus
Rutherford named it the nucleus; it is about 10⁻¹⁵ to 10⁻¹⁴ m across, while the atom is about 10⁻¹⁰ m.
!Common mistakeWriting 'proton' names one particle inside the nucleus, not the whole central core.
12Multiple choice · ★ Challenge
In a vacuum diode, the anode is made NEGATIVE with respect to the heated cathode. What is the current?
- ALarger than before, because both electrodes now emit electrons
- BThe same as before, because the cathode is still hot
- CZero: the emitted electrons are pushed back to the cathode
- DIt flows the other way, carried by positive ions
Show answer
Answer: C. Zero: the emitted electrons are pushed back to the cathode
Only the hot cathode emits electrons; a negative anode repels them, so none cross the tube. This one-way action lets a diode rectify a.c.
!Common mistakeChoosing 'both electrodes emit' forgets that the anode is cold; thermionic emission needs a hot surface.
13Multiple choice · ★ Challenge
Advertising signs filled with neon glow red, while signs filled with mercury vapour glow blue-green. What is the best explanation?
- AEach gas tube is made from a different coloured glass
- BNeon atoms are bigger, so they give out longer wavelengths
- CEach element has its own energy levels, so it emits its own wavelengths
- DThe two gases are at different temperatures, so they glow with different colours
Show answer
Answer: C. Each element has its own energy levels, so it emits its own wavelengths
The colours come from photons emitted when electrons fall between the energy levels of each element; different elements have different level spacings, giving different line spectra.
!Common mistakeChoosing 'different temperatures' describes the continuous glow of hot solids; in a discharge tube the colour depends on the element, not mainly on its temperature.
14Multiple choice
Which of these is NOT one of Bohr's postulates?
- AElectrons move only in certain allowed orbits
- BAn electron in an allowed orbit does not radiate energy
- CA photon is emitted only when an electron jumps to a lower level
- DElectrons give out energy continuously as they orbit the nucleus
Show answer
Answer: D. Electrons give out energy continuously as they orbit the nucleus
Bohr postulated the opposite: in an allowed orbit the electron does NOT radiate. Continuous radiation was the prediction of classical physics that made Rutherford's model unstable.
!Common mistakeChoosing 'an electron in an allowed orbit does not radiate' rejects Bohr's most important postulate.
15Multiple choice · ★ Challenge
Which observation could Bohr's model NOT explain?
- AHydrogen gas gives out light of only certain separate wavelengths
- BSome lines split into several close lines in a magnetic field
- CThe ionisation energy of hydrogen is 13.6 eV
- DHydrogen's Balmer series lies in the visible region
Show answer
Answer: B. Some lines split into several close lines in a magnetic field
Bohr's model predicts single, sharp levels, so it cannot explain the splitting of lines in a magnetic field (Zeeman effect) or their fine structure.
!Common mistakeChoosing the 13.6 eV ionisation energy picks one of Bohr's main successes; his model gives it exactly.
16Fill in the blank
In Bohr's model of hydrogen, the radius of the nth allowed orbit is proportional to ______.
Show answer
Answer: n²
rn = n²r₁, so the orbits are at 1, 4, 9, 16 … times the smallest radius.
!Common mistakeWriting 1/n² confuses the radius with the energy, which goes as −13.6/n² eV.
17Multiple choice · ★ Challenge
Why did Geiger and Marsden use an extremely thin gold foil?
- ASo most alpha particles get through, each meeting at most one nucleus
- BBecause gold is radioactive and adds to the alpha particles
- CBecause gold attracts alpha particles strongly
- DBecause a thick foil would reflect every one of the alpha particles straight back
Show answer
Answer: A. So most alpha particles get through, each meeting at most one nucleus
Gold can be beaten to a foil only a few hundred atoms thick, so alpha particles are not stopped and each large deflection comes from a single encounter with a nucleus.
!Common mistakeChoosing 'gold attracts alpha particles' is wrong: gold nuclei are positive and REPEL the positive alpha particles.
18Multiple choice
Which of these uses does NOT depend on atomic line spectra?
- AFinding which elements are present in the Sun's outer layers
- BMeasuring the mass of a planet from its moon's orbit
- CChecking a mineral sample for small amounts of metals
- DMaking street lamps that give a particular colour
Show answer
Answer: B. Measuring the mass of a planet from its moon's orbit
Planet masses come from gravity and orbit measurements (Unit 5); the other three all use the fact that each element emits or absorbs its own set of wavelengths.
!Common mistakeChoosing street lamps forgets that sodium and mercury lamps give their colours because of the line spectra of those elements.
19Short answer · ★ Challenge
State two successes and two failures of Bohr's model of the atom.
Show answer
Model answer: Successes (any two): explains why atoms are stable; explains line spectra; gives the correct wavelengths of the hydrogen series; gives the ionisation energy of hydrogen (13.6 eV). Failures (any two): does not work for atoms with more than one electron; cannot explain the relative brightness of lines; cannot explain the splitting of lines (fine structure, or in a magnetic field); mixes classical orbits with a quantum rule without justifying it.
!Common mistakeLearners often list 'cannot explain hydrogen' as a failure; hydrogen is exactly where Bohr's model works well.
20Multiple choice · ★ Challenge
Why was the alpha-scattering apparatus kept in an evacuated chamber?
- AAlpha particles can only be produced in a vacuum
- BA vacuum is needed so that the gold foil can conduct electricity
- CAir would stop or scatter the alpha particles within a few cm
- DAir would make the gold nuclei radioactive
Show answer
Answer: C. Air would stop or scatter the alpha particles within a few cm
Alpha particles lose energy quickly by ionising air; in a vacuum they reach the foil and the detector without being scattered by air molecules, so all deflections are caused by the foil.
!Common mistakeChoosing 'can only be produced in a vacuum' is wrong: radioactive sources emit alpha particles in air too, but the air absorbs them within a few centimetres.
21True or false
In the Franck–Hertz experiment, the current through mercury vapour drops sharply when the electrons reach about 4.9 eV, showing that mercury atoms take in energy only in fixed amounts.
Show answer
Answer: True
Electrons with less than 4.9 eV bounce elastically off mercury atoms; at 4.9 eV they can excite an atom, lose almost all their energy and fail to reach the collector, so the current falls.
!Common mistakeLearners often think atoms can absorb any small amount of energy; the sudden dips prove that the energy levels are discrete.
22Multiple choice · ★ Challenge
A mercury lamp gives out a bright green line of wavelength 546 nm. What is the energy of each photon in this line? (hc = 1240 eV nm)
- A0.440 eV
- B2.27 eV
- C3.64 × 10⁻¹⁹ eV
- D22.7 eV
Show answer
Answer: B. 2.27 eV
E = hc/λ = 1240 ÷ 546 = 2.27 eV.
!Common mistakeChoosing 3.64 × 10⁻¹⁹ eV gives the energy in JOULES but labels it eV; 3.64 × 10⁻¹⁹ J = 2.27 eV.
23True or false
Bohr's model explains why some lines in the hydrogen spectrum are brighter than others.
Show answer
Answer: False
Bohr's model gives the wavelengths of the lines but says nothing about how often each transition happens, so it cannot predict their brightness.
!Common mistakeLearners often think a model that gets the wavelengths right explains everything about the spectrum.
24Fill in the blank · ★ Challenge
In a Franck–Hertz experiment, dips in current occur at 4.9 V, 9.8 V and 14.7 V. The excited mercury atoms then emit light of wavelength ______ nm (hc = 1240 eV nm).
Show answer
Answer: 253
Each excitation needs 4.9 eV, so the emitted photons have 4.9 eV: λ = 1240 ÷ 4.9 = 253 nm (ultraviolet).
!Common mistakeWriting 84 nm (1240 ÷ 14.7) treats the third dip as one large excitation; it is three separate collisions of 4.9 eV each.
25Multiple choice
The glowing tungsten filament of an old-style lamp is viewed through a spectroscope. What kind of spectrum is seen?
- AA line spectrum characteristic of tungsten
- BAn absorption spectrum with dark lines
- CA single bright line of yellow light
- DA continuous spectrum of all colours
Show answer
Answer: D. A continuous spectrum of all colours
A hot, dense solid has atoms packed so closely that their levels merge into bands; it emits all wavelengths, giving a continuous spectrum.
!Common mistakeChoosing 'a line spectrum of tungsten' applies the rule for a hot low-pressure GAS to a solid; line spectra come from isolated atoms.
26Multiple choice · ★ Challenge
According to Bohr, what is the angular momentum of the electron in the n = 2 orbit of hydrogen? (h = 6.63 × 10⁻³⁴ J s)
- A1.06 × 10⁻³⁴ J s
- B2.11 × 10⁻³⁴ J s
- C4.22 × 10⁻³⁴ J s
- D1.33 × 10⁻³³ J s
Show answer
Answer: B. 2.11 × 10⁻³⁴ J s
mvr = nh/2π = 2 × 6.63 × 10⁻³⁴ ÷ 2π = 2.11 × 10⁻³⁴ J s.
!Common mistakeChoosing 1.33 × 10⁻³³ J s uses nh and forgets to divide by 2π.
27Short answer · ★ Challenge
Explain why the dips in current in the Franck–Hertz experiment with mercury are evenly spaced, about 4.9 V apart.
Show answer
Model answer: An electron that gains 4.9 eV can excite a mercury atom and lose that energy, then be accelerated again. With a 9.8 V accelerating p.d. it can gain 4.9 eV, lose it in a collision, gain another 4.9 eV and lose it again in a second collision, so the current dips again. Each extra 4.9 V allows one more inelastic collision, so the dips are spaced by the same 4.9 V.
!Common mistakeLearners often think the second dip means a second, higher energy level at 9.8 eV; it is the same 4.9 eV excitation happening twice.
28Fill in the blank
On an energy-level diagram, the level E = 0 (n = ∞) represents an electron that has been ______ from the atom.
Show answer
Answer: completely removed (freed)
At E = 0 the electron is just free of the nucleus; all bound levels are below zero, so they are negative.
!Common mistakeWriting 'in the ground state' confuses the top of the diagram with the bottom; the ground state is the most negative level.
29Multiple choice · ★ Challenge
Electrons are accelerated into hydrogen gas at low pressure, with all the atoms in the ground state. What is the smallest accelerating voltage at which the gas starts to give out the red Balmer line (n = 3 → n = 2)?
- A1.89 V
- B10.2 V
- C12.1 V
- D13.6 V
Show answer
Answer: C. 12.1 V
To emit the 3 → 2 line, atoms must first be raised from n = 1 to n = 3, which needs E₃ − E₁ = −1.51 − (−13.6) = 12.1 eV, so the electrons need 12.1 V.
!Common mistakeChoosing 1.89 V uses only the energy of the red photon; the atom must first get from the ground state all the way up to n = 3.
30Multiple choice
Why is the cathode of an electron gun (as in an old television tube or an X-ray tube) heated?
- ATo give electrons enough energy to escape
- BTo make the cathode glow so that the screen is lit
- CTo remove the air from the tube
- DTo make the cathode positively charged
Show answer
Answer: A. To give electrons enough energy to escape
Heating gives the conduction electrons enough thermal energy to overcome the work function: thermionic emission supplies the electron beam.
!Common mistakeChoosing 'to make the cathode glow' mistakes a side-effect for the purpose; the light from the cathode is not used.
31Short answer · ★ Challenge
Cool hydrogen gas absorbs certain wavelengths from white light, and the excited atoms soon re-emit photons of the same wavelengths. Explain why dark lines are still seen in the absorption spectrum.
Show answer
Model answer: The absorbed light comes from one direction (the beam towards the observer), but the excited atoms re-emit photons in all directions. So only a small fraction of the re-emitted light travels on in the original direction, and those wavelengths are much weaker than the rest of the spectrum: they appear as dark lines.
!Common mistakeLearners often think the absorbed energy disappears; it is re-emitted, but scattered in all directions.
32Multiple choice
In the modern (quantum) model of the atom, how is the electron in a hydrogen atom best described?
- AAs a tiny ball moving round a fixed circular orbit
- BAs a cloud showing where it is likely to be found
- CAs a particle at rest at a fixed point near the nucleus
- DAs positive charge spread through the whole atom
Show answer
Answer: B. As a cloud showing where it is likely to be found
Quantum mechanics gives only the probability of finding the electron at each place; the energy levels remain, but sharp circular orbits do not.
!Common mistakeChoosing 'a fixed circular orbit' describes Bohr's model, which the modern model replaced.
33Fill in the blank · ★ Challenge
An electron in a gas atom drops through an energy gap of 3.1 eV. The photon emitted has a frequency of ______ × 10¹⁴ Hz (h = 6.63 × 10⁻³⁴ J s, 1 eV = 1.6 × 10⁻¹⁹ J).
Show answer
Answer: 7.5
E = 3.1 × 1.6 × 10⁻¹⁹ = 4.96 × 10⁻¹⁹ J; f = E/h = 4.96 × 10⁻¹⁹ ÷ 6.63 × 10⁻³⁴ = 7.5 × 10¹⁴ Hz.
!Common mistakeWriting 4.7 × 10³³ Hz (3.1 ÷ 6.63 × 10⁻³⁴) forgets to change eV into joules before dividing by h.
34Fill in the blank · ★ Challenge
Hydrogen atoms falling from n = 5 to n = 3 give the second line of the Paschen series. Found with the Rydberg formula (R = 1.097 × 10⁷ m⁻¹), its wavelength is about ______ nm.
Show answer
Answer: 1280
1/λ = 1.097 × 10⁷ × (1/9 − 1/25) = 1.097 × 10⁷ × 0.0711 = 7.80 × 10⁵ m⁻¹, so λ = 1.28 × 10⁻⁶ m = 1280 nm (infrared).
!Common mistakeWriting 1880 nm gives the FIRST Paschen line (4 → 3); the second line comes from n = 5.
35Fill in the blank
An alpha particle carries a charge of +2e. When it is accelerated from rest through a potential difference of 1000 V, it gains ______ eV of kinetic energy.
Show answer
Answer: 2000
Energy gained = qV = 2e × 1000 V = 2000 eV.
!Common mistakeWriting 1000 eV treats the alpha particle as if it had the charge of one electron; the energy in eV equals the charge (in units of e) times the p.d.
36Short answer · ★ Challenge
Explain why the Balmer series of hydrogen is in the visible region while the Lyman series is in the ultraviolet.
Show answer
Model answer: The Lyman series consists of jumps down to n = 1. The gap between n = 1 and the other levels is large (at least 10.2 eV), so the photons have high energy and short wavelengths (122 nm or less): ultraviolet. Balmer jumps end on n = 2, where the gaps are much smaller (1.9 to 3.4 eV), giving wavelengths of about 365–656 nm, mostly in the visible range.
!Common mistakeLearners often say the Balmer series is visible because the electron 'starts from a higher level'; what matters is the level it ends on and the size of the energy gap.
37Short answer
The filaments of many electron tubes are coated with barium or strontium oxide. Explain why.
Show answer
Model answer: These oxides have a low work function, so less energy is needed for an electron to escape from the surface. More electrons are released by thermionic emission at a given temperature, or the same emission is obtained at a lower temperature, which saves power and makes the filament last longer.
!Common mistakeLearners often say the coating 'conducts better'; the purpose is to lower the work function, not the resistance.
38Multiple choice · ★ Challenge
A line in the Balmer series of hydrogen has a wavelength of 410 nm. Using 1/λ = R(1/2² − 1/n²) with R = 1.097 × 10⁷ m⁻¹, which transition produces it?
- An = 6 → n = 2
- Bn = 5 → n = 2
- Cn = 7 → n = 2
- Dn = 4 → n = 2
Show answer
Answer: A. n = 6 → n = 2
1/(410 × 10⁻⁹ × 1.097 × 10⁷) = 0.222 = 1/4 − 1/n², so 1/n² = 0.028 and n² = 36: n = 6.
!Common mistakeChoosing n = 5 → 2 gives 434 nm; check your answer by putting n back into the formula.
39Multiple choice
In Bohr's model the radius of the smallest hydrogen orbit (n = 1) is 0.053 nm. What is the radius of the n = 3 orbit?
- A0.16 nm
- B0.018 nm
- C0.48 nm
- D0.0059 nm
Show answer
Answer: C. 0.48 nm
r ∝ n²: r₃ = 3² × 0.053 = 9 × 0.053 = 0.48 nm.
!Common mistakeChoosing 0.16 nm multiplies by n instead of n²; the radius grows with the SQUARE of the orbit number.
40Short answer · ★ Challenge
A gold atom has a radius of about 1.4 × 10⁻¹⁰ m and its nucleus a radius of about 7 × 10⁻¹⁵ m. If the nucleus were enlarged to the size of a 1 cm marble, how wide would the whole atom be? What does this tell you about the atom?
Show answer
Model answer: Ratio of sizes = 1.4 × 10⁻¹⁰ ÷ 7 × 10⁻¹⁵ = 2 × 10⁴. A 1 cm nucleus would sit inside an atom 2 × 10⁴ × 1 cm = 200 m across, about the size of a large stadium. The atom is almost entirely empty space, with nearly all its mass in the tiny nucleus.
!Common mistakeLearners often cube the ratio, as if comparing volumes; the question asks for widths, which scale by the ratio of the radii only.
41Multiple choice · ★ Challenge
How much energy, in joules, is needed to ionise a hydrogen atom whose electron is in the n = 3 level (E₃ = −1.51 eV; 1 eV = 1.6 × 10⁻¹⁹ J)?
- A2.4 × 10⁻¹⁹ J
- B1.9 × 10⁻¹⁸ J
- C2.2 × 10⁻¹⁸ J
- D9.4 × 10¹⁸ J
Show answer
Answer: A. 2.4 × 10⁻¹⁹ J
To ionise, the electron must be raised from −1.51 eV to 0: 1.51 eV = 1.51 × 1.6 × 10⁻¹⁹ = 2.4 × 10⁻¹⁹ J.
!Common mistakeChoosing 2.2 × 10⁻¹⁸ J uses 13.6 eV, the ionisation energy from the GROUND state; an excited atom needs much less.
42Short answer
Thomson found that cathode rays are made of tiny negative particles (electrons) that come from all kinds of atoms. Explain why he concluded that atoms must also contain positive charge, and why most of an atom's mass could not be in its electrons.
Show answer
Model answer: Atoms are electrically neutral, so the negative charge of the electrons must be balanced by an equal amount of positive charge. Electrons are about 2000 times lighter than even the lightest atom (hydrogen), so they can make up only a tiny fraction of the atom's mass; most of the mass must be in the positive part.
!Common mistakeLearners often think atoms are made only of electrons; that would give every atom a negative charge.
43Multiple choice · ★ Challenge
Electron A is accelerated from rest through 100 V and electron B through 400 V. How does B's final speed compare with A's?
- A4 times larger
- B16 times larger
- C√2 times larger
- D2 times larger
Show answer
Answer: D. 2 times larger
eV = ½mv², so v ∝ √V: vB/vA = √(400 ÷ 100) = √4 = 2.
!Common mistakeChoosing 4 times assumes the speed is proportional to the p.d.; it is the kinetic energy that is proportional to V, and v ∝ √(KE).
44Short answer · ★ Challenge
A learner draws the hydrogen energy levels as +13.6 eV (n = 1), +3.4 eV (n = 2) and +1.51 eV (n = 3). Explain what is wrong, and correct the diagram.
Show answer
Model answer: The energies should be negative: −13.6, −3.40 and −1.51 eV. Zero energy is chosen for a free electron at rest far away (n = ∞), and a bound electron has less energy than a free one, so all bound levels are below zero. In the learner's diagram the ground state (n = 1) would have the HIGHEST energy, which is the wrong way round.
!Common mistakeLearners often drop the minus signs because 'energy cannot be negative'; the sign comes from the choice of zero at infinity.
45Short answer
Explain the stability problem of Rutherford's nuclear atom and how Bohr's postulates solved it.
Show answer
Model answer: In Rutherford's model the electron moves in a circle, so it is accelerating; classical physics says an accelerating charge radiates energy, so the electron would spiral into the nucleus in a tiny fraction of a second and atoms could not be stable. Bohr postulated that electrons can occupy only certain allowed orbits in which they do not radiate, and that energy is emitted only as a photon when an electron jumps between orbits. So an electron in the lowest orbit cannot lose any more energy and the atom is stable.
!Common mistakeLearners often say the electron stays up because it moves fast; speed alone does not stop a classically radiating charge from spiralling in.
46Short answer · ★ Challenge
An atom has energy levels at −8.0 eV (ground state), −5.0 eV and −3.0 eV. List all the photon energies it can emit, find each wavelength (hc = 1240 eV nm), and say which are visible (400–700 nm).
Show answer
Model answer: Possible jumps: −3.0 → −5.0 gives 2.0 eV, λ = 1240 ÷ 2.0 = 620 nm (visible, orange-red); −5.0 → −8.0 gives 3.0 eV, λ = 413 nm (visible, violet); −3.0 → −8.0 gives 5.0 eV, λ = 248 nm (ultraviolet, not visible). So two of the three lines are visible.
!Common mistakeLearners often forget the direct jump from the top level to the ground state; with three levels there are three possible lines.
47Multiple choice
An electron is accelerated from rest through a potential difference of 5000 V. How much kinetic energy does it gain?
- A8.0 × 10⁻¹⁶ J
- B5000 J
- C3.1 × 10²² J
- D3.2 × 10⁻²³ J
Show answer
Answer: A. 8.0 × 10⁻¹⁶ J
KE = eV = 1.6 × 10⁻¹⁹ × 5000 = 8.0 × 10⁻¹⁶ J (which is 5000 eV).
!Common mistakeChoosing 5000 J mixes up joules with electronvolts: the energy is 5000 eV, and 1 eV is only 1.6 × 10⁻¹⁹ J.
48Short answer · ★ Challenge
In a small research accelerator, protons (mass 1.67 × 10⁻²⁷ kg, charge 1.6 × 10⁻¹⁹ C) are accelerated from rest through 20 kV. Find their kinetic energy in eV and in joules, and their final speed.
Show answer
Model answer: KE = 20 000 eV = 2.0 × 10⁴ × 1.6 × 10⁻¹⁹ = 3.2 × 10⁻¹⁵ J. ½mv² = KE, so v = √(2 × 3.2 × 10⁻¹⁵ ÷ 1.67 × 10⁻²⁷) = √(3.83 × 10¹²) = 2.0 × 10⁶ m/s.
!Common mistakeLearners often use the electron's mass for a proton, which gives a speed √1836 ≈ 43 times too large.
49Short answer
An electron with 11 eV of kinetic energy can excite a ground-state hydrogen atom to n = 2 (needing 10.2 eV), but a photon of 11 eV cannot. Explain the difference.
Show answer
Model answer: A photon is absorbed whole or not at all, so its energy must exactly match a gap between levels; 11 eV matches no gap, so the photon passes through. An electron can hand over just part of its energy in a collision: it gives 10.2 eV to the atom and moves on with the remaining 0.8 eV.
!Common mistakeLearners often think photons and electrons excite atoms in the same way; only a photon needs an exact energy match.
50Short answer · ★ Challenge
Use Bohr's condition mvr = nh/2π to find the speed of the electron in the ground state of hydrogen (r₁ = 5.3 × 10⁻¹¹ m, m = 9.11 × 10⁻³¹ kg, h = 6.63 × 10⁻³⁴ J s). What fraction of the speed of light is this?
Show answer
Model answer: For n = 1: v = h/(2πmr) = 6.63 × 10⁻³⁴ ÷ (2π × 9.11 × 10⁻³¹ × 5.3 × 10⁻¹¹) = 2.2 × 10⁶ m/s. v/c = 2.2 × 10⁶ ÷ 3.0 × 10⁸ = 0.0073, less than 1% of the speed of light.
!Common mistakeLearners often forget the 2π in nh/2π, which gives a speed 2π times too large.