E.1 Structure of the atom – IB Physics exam questions with mark scheme

1Core3 marks

A nucleus of uranium-235 is represented by ²³⁵₉₂U.

(a) State the number of protons and the number of neutrons in this nucleus. [2]

(b) Define the term isotopes. [1]

(a) 92 protons(1)
(a) 235 − 92 = 143 neutrons(1)
(b) Nuclei/atoms of the same element with the same number of protons (Z) but a different number of neutrons (accept: same proton/atomic number, different nucleon/mass number; do not accept “different number of protons” or “different mass” alone)(1)
Answer: (a) 92 protons, 143 neutrons; (b) same Z, different N
Examiner tip: Neutrons = A − Z; isotopes differ in neutrons, not in protons.

2Core4 marks

The energy of the n^th level of hydrogen is E_n = −13.6/n² eV.

(a) Explain why the emission spectrum of hydrogen consists of discrete lines. [2]

(b) Calculate the wavelength of the photon emitted when an electron falls from n = 3 to n = 2. [2]

(a) Electrons can only occupy discrete (fixed) energy levels(1)
(a) A photon is emitted when an electron drops between levels, with photon energy = difference in level energies, so only specific frequencies/wavelengths appear(1)
(b) ΔE = −1.51 − (−3.40) = 1.89 eV = 3.02 × 10⁻¹⁹ J(1)
(b) λ = hc/ΔE = 6.58 × 10⁻⁷ m (658 nm; allow 6.56–6.58 × 10⁻⁷ m)(1)
Answer: (a) discrete levels, photon energy = level difference; (b) 658 nm
Examiner tip: Convert eV to J before using λ = hc/E; say the photon energy equals the level difference.

3Core4 marks

An iron ion is represented by $^{56}_{26}\text{Fe}^{3+}$.

(a) State the number of protons, neutrons and electrons in this ion. [2]

(b) Calculate the charge on the nucleus of this ion. [1]

$^{40}_{18}\text{Ar}\qquad ^{40}_{19}\text{K}\qquad ^{39}_{19}\text{K}\qquad ^{40}_{20}\text{Ca}$

Identify the two nuclides that are isotopes of the same element. [1]

(a) 26 protons and 30 neutrons(1)
(a) 23 electrons (do not accept 26 or 29)(1)
(b) $26\times1.60\times10^{-19}=4.2\times10^{-18}\ \text{C}$ (accept $4.16\times10^{-18}$ C; do not accept a charge based on 23 or 3 charges: the nucleus charge is set by the protons only)(1)
(c) $^{40}_{19}\text{K}$ and $^{39}_{19}\text{K}$ (both required; same proton number, different nucleon number)(1)
Answer: (a) 26 p, 30 n, 23 e⁻; (b) $4.2\times10^{-18}$ C; (c) K-40 and K-39
Examiner tip: The ion's 3+ charge changes only the electron count; students often use 23 or 3 in part (b), but the nucleus holds all 26 protons.

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4Standard6 marks

The diagram shows the four lowest energy levels of an isolated atom of element X. The energies are −9.2 eV, −4.6 eV, −2.9 eV and −1.5 eV.

(a) State the number of different photon energies that could be emitted when the atoms are excited to the −1.5 eV level and then return to the ground state. [1]

(b) Calculate the longest wavelength of photon that can be emitted in transitions between these levels. [2]

(c) A beam of photons with energies 4.6 eV, 5.0 eV and 6.3 eV passes through a sample of atoms of X in the ground state. Identify which photon energies are absorbed, and explain your answer. [2]

(d) Determine, in J, the ionisation energy of an atom of X in the ground state. [1]

(a) 6(1)
(b) smallest energy gap identified: $\Delta E=-1.5-(-2.9)=1.4\ \text{eV}$ (and converted: $1.4\times1.60\times10^{-19}=2.24\times10^{-19}$ J)(1)
(b) $\lambda=\dfrac{hc}{\Delta E}=\dfrac{6.63\times10^{-34}\times3.00\times10^{8}}{2.24\times10^{-19}}=8.9\times10^{-7}\ \text{m}$ (accept 890 nm or 888 nm; award [1 max] for using the 1.7 eV gap, giving 731 nm)(1)
(c) 4.6 eV and 6.3 eV absorbed; 5.0 eV not absorbed(1)
(c) photon is absorbed only if its energy equals exactly the difference between the ground state and a higher level (−9.2 to −4.6 = 4.6 eV, −9.2 to −2.9 = 6.3 eV); no level is 5.0 eV above the ground state(1)
(d) $9.2\times1.60\times10^{-19}=1.5\times10^{-18}\ \text{J}$ (accept $1.47\times10^{-18}$ J)(1)
Answer: (a) 6; (b) $8.9\times10^{-7}$ m; (c) 4.6 eV and 6.3 eV; (d) $1.5\times10^{-18}$ J
Examiner tip: Longest wavelength means the smallest energy gap, not the biggest; and an atom cannot absorb part of a photon's energy, so 5.0 eV passes straight through.

5Standard6 marks

Light from a star is passed through a diffraction grating. Its spectrum is continuous apart from a number of dark lines. The diagram shows the wavelengths, in nm, of the dark lines in the visible region.

The table gives the strongest visible lines of four elements, measured in the laboratory.

Element	Wavelengths of lines / nm
hydrogen	434.0, 486.1, 656.3
helium	447.1, 501.6, 587.6
sodium	589.0, 589.6
magnesium	516.7, 517.3, 518.4

(a) Identify, with a reason, the elements present in the outer layers of the star. [2]

(b) Explain why the lines appear dark against the continuous spectrum. [2]

(c) Calculate, in eV, the energy difference between the two atomic energy levels responsible for the 589.0 nm line. [2]

(a) hydrogen and sodium (both needed; do not credit if helium or magnesium also stated)(1)
(a) every laboratory line of these elements matches a dark line / helium (587.6, 447.1, 501.6 nm) and magnesium (516.7–518.4 nm) lines are missing from the star's spectrum(1)
(b) atoms in the cooler outer gas absorb photons whose energy equals the difference between two of their energy levels(1)
(b) the excited atoms re-emit these photons in random directions, so fewer of those wavelengths reach the observer (accept 'scattered out of the beam')(1)
(c) $E=\dfrac{hc}{\lambda}=\dfrac{6.63\times10^{-34}\times3.00\times10^{8}}{589.0\times10^{-9}}=3.38\times10^{-19}\ \text{J}$(1)
(c) $\dfrac{3.38\times10^{-19}}{1.60\times10^{-19}}=2.1\ \text{eV}$ (accept 2.11 eV)(1)
Answer: (a) hydrogen and sodium; (b) absorption then re-emission in all directions; (c) 2.1 eV
Examiner tip: Each element's lines are a fingerprint: an element is only present if all its strong lines appear, so helium's 587.6 nm near the sodium pair is a trap.

6StandardHL only6 marks

The radius $R$ of a nucleus with nucleon number $A$ is given by $R=R_0A^{1/3}$, where $R_0=1.20\times10^{-15}$ m.

(a) Calculate the radius of a nucleus of $^{56}_{26}\text{Fe}$. [1]

(b) Show that the density of nuclear matter is independent of $A$, and calculate its value. Take the mass of a nucleon to be $1.66\times10^{-27}$ kg. [3]

(c) A nucleus Y has a radius exactly twice that of a nucleus of $^{27}_{13}\text{Al}$. Determine the nucleon number of Y. [2]

(a) $R=1.20\times10^{-15}\times56^{1/3}=4.6\times10^{-15}\ \text{m}$ (accept 4.59 fm)(1)
(b) $\rho=\dfrac{Am_{\text{n}}}{\frac{4}{3}\pi R^{3}}$ with $R^{3}=R_0^{3}A$ substituted(1)
(b) $A$ cancels, giving $\rho=\dfrac{3m_{\text{n}}}{4\pi R_0^{3}}$, which contains only constants(1)
(b) $\rho=2.3\times10^{17}\ \text{kg m}^{-3}$ (accept $2.29\times10^{17}$; a value calculated for Fe-56 alone scores this mark but not the 'independent' marks)(1)
(c) $A\propto R^{3}$ so $A_Y=27\times2^{3}$(1)
(c) $A_Y=216$(1)
Answer: (a) $4.6\times10^{-15}$ m; (b) $2.3\times10^{17}$ kg m⁻³; (c) 216
Examiner tip: In (c), doubling the radius multiplies the volume, and therefore the nucleon number, by eight, not two.

7HardHL only7 marks

In a scattering experiment, alpha particles of kinetic energy 5.5 MeV are fired at a thin gold foil. Gold has proton number 79 and nucleon number 197. Assume the gold nucleus does not recoil.

(a) Calculate the distance of closest approach of an alpha particle that approaches a gold nucleus head-on. [3]

(b) Using $R=R_0A^{1/3}$ with $R_0=1.20\times10^{-15}$ m, compare your answer to (a) with the radius of the gold nucleus. [1]

(c) The experiment is repeated with alpha particles of much higher energy. Above a certain energy the number scattered through large angles is less than the Rutherford model predicts. Estimate this energy, in MeV, by assuming the alpha particle (A = 4) just touches the surface of the gold nucleus. [2]

(d) Suggest why the scattering deviates from the Rutherford prediction above this energy. [1]

(a) at closest approach all kinetic energy becomes electric potential energy: $E_k=\dfrac{k(2e)(79e)}{d}$(1)
(a) $E_k=5.5\times10^{6}\times1.60\times10^{-19}=8.8\times10^{-13}$ J(1)
(a) $d=\dfrac{8.99\times10^{9}\times158\times(1.60\times10^{-19})^{2}}{8.8\times10^{-13}}=4.1\times10^{-14}\ \text{m}$(1)
(b) $R_{\text{Au}}=1.20\times10^{-15}\times197^{1/3}=7.0\times10^{-15}$ m, so $d$ is about six times larger / the alpha particle does not reach the nucleus (allow ECF from (a))(1)
(c) separation $=R_0(197^{1/3}+4^{1/3})=8.9\times10^{-15}$ m(1)
(c) $E=\dfrac{k(158)e^{2}}{8.9\times10^{-15}}=4.1\times10^{-12}\ \text{J}\approx26\ \text{MeV}$ (accept 25–26 MeV; allow ~33 MeV, i.e. [1 max], if the alpha radius is ignored)(1)
(d) the alpha particle comes within range of the (short-range, attractive) strong nuclear force, so the force is no longer purely Coulomb (accept 'the alpha particle enters / interacts with the nucleus')(1)
Answer: (a) $4.1\times10^{-14}$ m; (b) about 6 × the gold radius of $7.0\times10^{-15}$ m; (c) ≈ 26 MeV; (d) strong nuclear force acts
Examiner tip: Use a charge of 2e for the alpha particle and 79e for gold, and convert MeV to joules before dividing; forgetting the factor of 2 halves the answer.

8StretchHL only8 marks

In the Bohr model of hydrogen, an electron of mass $m_e$ moves in a circular orbit of radius $r$ around a proton. The angular momentum of the electron is quantised according to $m_evr=\dfrac{nh}{2\pi}$.

(a) Show that the radius of the $n$th orbit is $$r_n=\frac{n^{2}h^{2}}{4\pi^{2}km_ee^{2}}$$ where $k$ is the Coulomb constant. [3]

(b) Calculate the radius of the orbit for $n=1$. [1]

(d) Show that the total energy of the electron in the $n=1$ orbit is approximately −13.6 eV. [2]

(a) electric force provides the centripetal force: $\dfrac{ke^{2}}{r^{2}}=\dfrac{m_ev^{2}}{r}$(1)
(a) from quantisation: $v=\dfrac{nh}{2\pi m_er}$(1)
(a) substitution: $\dfrac{ke^{2}}{r}=m_e\dfrac{n^{2}h^{2}}{4\pi^{2}m_e^{2}r^{2}}$, rearranged to the given result (an answer that does not show the substitution step loses this mark)(1)
(b) $r_1=\dfrac{(6.63\times10^{-34})^{2}}{4\pi^{2}\times8.99\times10^{9}\times9.11\times10^{-31}\times(1.60\times10^{-19})^{2}}=5.3\times10^{-11}\ \text{m}$(1)
(c) $v=\dfrac{h}{2\pi m_er_1}$ (or from $v^{2}=\dfrac{ke^{2}}{m_er}$)(1)
(c) $v=2.2\times10^{6}\ \text{m s}^{-1}$ (allow ECF from (b))(1)
(d) $E_k=\tfrac{1}{2}m_ev^{2}=2.17\times10^{-18}$ J (13.5 eV) and $E_p=-\dfrac{ke^{2}}{r}=-4.33\times10^{-18}$ J (−27.1 eV)(1)
(d) $E=E_k+E_p=-2.17\times10^{-18}\ \text{J}\approx-13.5\ \text{eV}$, i.e. approximately −13.6 eV (accept use of $E=-\dfrac{ke^{2}}{2r}$ derived from $E_k=-\tfrac{1}{2}E_p$)(1)
Answer: (b) $5.3\times10^{-11}$ m; (c) $2.2\times10^{6}$ m s⁻¹; (d) $E\approx-2.2\times10^{-18}$ J ≈ −13.5 eV
Examiner tip: In a 'show that' give the calculated value to one more significant figure than the target (−13.54 eV here); just writing −13.6 eV earns nothing.

9HardHL only6 marks

The energy levels of the hydrogen atom are given by $E_n=-\dfrac{13.6}{n^{2}}$ eV.

(a) A hydrogen atom in its ground state absorbs a photon of wavelength 102.6 nm. Determine the value of $n$ of the level to which the electron is excited. [3]

(b) A photon of energy 11.0 eV is not absorbed by a hydrogen atom in its ground state, but an electron with kinetic energy 11.0 eV can excite the atom. Explain this difference. [2]

(c) According to the Bohr model, the orbit radius is proportional to $n^{2}$. State the ratio of the orbit radius for the level in (a) to the ground-state radius. [1]

(a) $E=\dfrac{hc}{\lambda}=\dfrac{6.63\times10^{-34}\times3.00\times10^{8}}{102.6\times10^{-9}}=1.94\times10^{-18}\ \text{J}=12.1\ \text{eV}$(1)
(a) new level energy $=-13.6+12.1=-1.5$ eV(1)
(a) $n^{2}=\dfrac{13.6}{1.5}\approx9$ so $n=3$ (accept answers that compare 12.1 eV with $13.6(1-1/n^{2})$ for n = 2, 3, 4)(1)
(b) a photon is absorbed only if its whole energy equals a level difference (10.2 eV or 12.1 eV), and 11.0 eV matches no difference(1)
(b) an electron can transfer just part of its kinetic energy in a collision (e.g. 10.2 eV to reach n = 2), keeping the rest (0.8 eV) as kinetic energy(1)
(c) 9 (allow ECF from (a), i.e. $n^{2}$)(1)
Answer: (a) $n=3$; (b) photon must give all its energy; electron can give part; (c) 9
Examiner tip: Photons are all-or-nothing, colliding electrons are not; say both halves of that contrast to get both marks in (b).

10Hard6 marks

In the Geiger–Marsden experiment, a narrow beam of alpha particles was directed at a very thin gold foil in a vacuum, and the scattered alpha particles were detected at different angles.

Describe the main observations of the experiment and explain how each led Rutherford to the nuclear model of the atom. [6]

Observation: almost all alpha particles passed straight through the foil, or were deflected through very small angles(1)
Conclusion: most of the atom is empty space / the positive charge and mass are not spread through the whole atom (contradicts the 'plum-pudding' model)(1)
Observation: a small fraction were deflected through large angles(1)
Observation: a very small fraction (about 1 in 8000) were deflected through more than 90°, i.e. bounced back(1)
Conclusion: back-scattering requires a very strong repulsive electric force, so the positive charge is concentrated in a very small volume(1)
Conclusion: since alpha particles are heavy and fast, the nucleus must also contain most of the atom's mass(1)
Conclusion: the rarity of large deflections shows the nucleus is tiny compared with the atom (radius about $10^{-15}$ m versus $10^{-10}$ m)(1)
Model: a small, dense, positively charged nucleus with electrons occupying the space around it(1)
Experimental detail: vacuum so alphas are not stopped by air; thin foil so each alpha is scattered by a single nucleus(1)

Level	Marks	What it looks like
Level 3	5–6	Describes all three observations (most undeflected, some large-angle, a few back-scattered) and links each clearly to a feature of the nuclear model: empty space, small concentrated positive charge, and concentrated mass. The account is logically ordered and uses correct terminology.
Level 2	3–4	Describes at least two observations and links at least two of them to features of the model, though links may be incomplete or one feature (e.g. mass) is missing.
Level 1	1–2	Gives one or more relevant observations or features of the model, with little or no linking between them.
Level 0	0	No relevant content.

Answer: Most pass straight through (atom mostly empty); a few deflected through large angles and some reflected (tiny, massive, positive nucleus); model: small dense positive nucleus with electrons outside.
Examiner tip: Students lose marks by listing observations without linking each one to a specific feature of the model; pair every observation with its conclusion.

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