IB Physics E.1: Structure of the atom worked solutions

The lower number is the proton number: 26. Neutrons = 56 - 26 = 30. The neutral atom has 26 electrons and the 3+ ion has lost three, leaving 23. B adds three electrons instead of removing them; C changes the proton number, which would make a different element; D reads the nucleon number as the neutron number.

An atom is about 10^{-10}\,\mathrm{m} across and a nucleus about 10^{-15}\,\mathrm{m}, a ratio of 10^{5}. 2\,\mathrm{mm}\times10^{5} = 2\times10^{2}\,\mathrm{m} = 200\,\mathrm{m}. A, B and D use ratios of 10^{3}, 10^{4} and 10^{7}.

Turning an alpha particle back needs a very large repulsive force, which only a small, concentrated positive charge can supply. A is a real observation, but it shows that the atom is mostly empty space. C was not observed. D is wrong: the nucleus and the alpha particle are both positive, so the force is repulsive.

A line spectrum comes from electrons dropping between fixed atomic energy levels; each drop releases one photon of energy hf equal to the gap. A describes gamma emission from nuclei. B is false: many lines are in the ultraviolet or infrared. D is the opposite of what the lines show: only certain energies are allowed.

Six downward jumps are possible, with gaps 2.0, 2.0, 4.0, 4.0, 6.0 and 8.0\,\mathrm{eV}. Equal gaps give the same photon energy and so the same wavelength: only four values, 2.0, 4.0, 6.0 and 8.0\,\mathrm{eV}. C counts jumps, not distinct gaps; A counts only jumps to the lowest level (it has three).

Longest wavelength means smallest photon energy, so the smallest downward gap: E_4\to E_3, 0.6\,\mathrm{eV}. A is the largest gap (shortest wavelength). D has the same gap but goes up, so it is an absorption, not an emission. B is 1.5\,\mathrm{eV}.

E = 2.6\times1.60\times10^{-19} = 4.16\times10^{-19}\,\mathrm{J}; \lambda = hc/E = 1.989\times10^{-25}/4.16\times10^{-19} = 4.8\times10^{-7}\,\mathrm{m}, which is 478\,\mathrm{nm}, inside the visible range 400–700\,\mathrm{nm}. C forgets to convert eV to J; D slips a factor of 100 converting to nanometres; B names the wrong region.

Absorption needs a photon whose energy matches, exactly, a gap measured up from the ground state: 4.0, 5.5 or 8.0\,\mathrm{eV}. Only 4.0\,\mathrm{eV} is offered. A is the value of a level, not a gap. C is the gap between -5.0 and -3.5\,\mathrm{eV}, but no atom starts at -5.0\,\mathrm{eV}. D would lift the electron to -3.0\,\mathrm{eV}, which is not a level.

Bohr allowed only orbits with mvr = nh/2\pi, and in those orbits the electron does not radiate. A is the classical prediction Bohr rejected. C is the radioactive decay law (E.3). D is the classical picture; Bohr's condition restricts the radius to r\propto n^2.

L = nh/2\pi, so L\propto n and the ratio is 3/1 = 3. C uses the radius ratio n^2; A and D invert.

R = R_0A^{1/3}, so A\propto R^3. Doubling R multiplies A by 2^3 = 8: 8\times27 = 216. A doubles A; B multiplies by 4 (R^2); D multiplies the proton number 13 by 8.

The model needs exactly one electron. Electrons = Z - charge: A has 3-1 = 2; B has 3-2 = 1; C is a neutral atom with 2; D has 4-2 = 2. The nucleon number plays no part in the count.

(a) Nucleon number 79 + 118 = 197 on top, proton number 79 below: ^{197}_{79}\mathrm{Au} (A1). (b) Nearly every alpha particle goes through undeviated or almost so (A1): the atom must be mostly empty (A1). A tiny minority swing through large angles, a few beyond 90^\circ (A1); that needs a strong repulsion from a tiny region holding the atom's positive charge and almost all its mass, the nucleus (A1). (c) As the alpha particle approaches, the repulsion does negative work on it: its kinetic energy decreases and the electric potential energy increases by the same amount, until the alpha particle stops for an instant (M1). On the way out the potential energy is converted back, and it leaves along the same line with its original kinetic energy (A1).

(a) The hydrogen atoms have discrete energy levels, so they absorb only photons whose energy hf exactly equals the difference between two levels (M1). Absorbing a photon lifts an electron to a higher level (A1). The atoms then de-excite and re-emit the energy in all directions, so very little returns to the original beam and those wavelengths appear dark (A1). (b) E = hc/\lambda = (6.63\times10^{-34}\times3.00\times10^{8})/(434\times10^{-9}) = 4.58\times10^{-19}\,\mathrm{J} (M1); \div\,1.60\times10^{-19} = 2.86\,\mathrm{eV} (A1). (c) Each element has its own set of lines, so the matching pattern shows that hydrogen is present in the cooler gas around the star (A1).

(a) Zero is the energy of an electron that has just escaped the atom; a negative value means the electron is bound and that much energy must be supplied to free it (A1). (b) Four levels give 4\times3/2 = 6 downward jumps, with gaps 1.0, 2.2, 3.2, 7.2, 9.4 and 10.4\,\mathrm{eV}, all different: 6 wavelengths (A1). (c) Longest from the smallest gap, 1.0\,\mathrm{eV}: \lambda = hc/E = 1.989\times10^{-25}/1.60\times10^{-19} = 1.24\times10^{-6}\,\mathrm{m} (M1)(A1). Shortest from 10.4\,\mathrm{eV}: 1.989\times10^{-25}/(10.4\times1.60\times10^{-19}) = 1.20\times10^{-7}\,\mathrm{m} (A1). (d) Visible photons have energies between hc/700\,\mathrm{nm} = 1.78\,\mathrm{eV} and hc/400\,\mathrm{nm} = 3.11\,\mathrm{eV} (M1). Only the 2.2\,\mathrm{eV} gap lies in this band: \lambda = 565\,\mathrm{nm}. The 3.2\,\mathrm{eV} line is at 388\,\mathrm{nm}, just into the ultraviolet (A1).

(a) E_{CB} = hc/\lambda = 1.989\times10^{-25}/820\times10^{-9} = 2.43\times10^{-19}\,\mathrm{J} = 1.52\,\mathrm{eV} (M1). E_{BA} = 1.989\times10^{-25}/410\times10^{-9} = 4.85\times10^{-19}\,\mathrm{J} = 3.03\,\mathrm{eV} (A1). The gaps add: E_{CA} = 1.52 + 3.03 = 4.55\,\mathrm{eV} (A1). (b) \lambda = hc/E_{CA} = 1.989\times10^{-25}/(4.55\times1.60\times10^{-19}) = 273\,\mathrm{nm}; the same comes from 1/\lambda = 1/820 + 1/410 (A1). (c) Photon energies add because the levels are fixed; since E\propto1/\lambda, the direct jump has the largest energy and so the SHORTEST wavelength, not a longer one (A1).

(a) The largest photon energy spans the whole diagram (M1): highest level = -6.0 + 4.4 = -1.6\,\mathrm{eV} (A1). The 1.8\,\mathrm{eV} photon goes from the highest level to the middle one: -1.6 - 1.8 = -3.4\,\mathrm{eV}; check: -3.4 - (-6.0) = 2.6\,\mathrm{eV}, the third photon (A1). (b) An atom absorbs a photon only if its energy exactly equals a gap that starts at the level the atom is in (M1). From -6.0\,\mathrm{eV} the gaps are 2.6\,\mathrm{eV} and 4.4\,\mathrm{eV}, so these are absorbed (A1). The 1.8\,\mathrm{eV} gap starts at the middle level, which no atom in a cool sample occupies, so these photons pass through (A1).

(a) Gaps: 5.0, 4.8 and 0.2\,\mathrm{eV} (M1). \lambda = hc/E: 1.989\times10^{-25}/(5.0\times1.60\times10^{-19}) = 249\,\mathrm{nm} and 1.989\times10^{-25}/(4.8\times1.60\times10^{-19}) = 259\,\mathrm{nm} (A1); 1.989\times10^{-25}/(0.2\times1.60\times10^{-19}) = 6.22\times10^{-6}\,\mathrm{m}, in the infrared (A1). (b) The two lowest levels are only 0.2\,\mathrm{eV} apart, so the two jumps from E_3 differ in energy, and so in wavelength, by only a small amount (A1). (c) Two: the infrared line is far outside the range (A1).

(a) The electric force provides the centripetal force: ke^2/r^2 = mv^2/r (M1), so v = \sqrt{ke^2/(mr)} = \sqrt{8.99\times10^{9}\times(1.60\times10^{-19})^2/(9.11\times10^{-31}\times5.3\times10^{-11})} = 2.18\times10^{6}\,\mathrm{m\,s^{-1}} (A1). (b) E_K = \tfrac12mv^2 = 2.17\times10^{-18}\,\mathrm{J} (M1) = 13.6\,\mathrm{eV} (A1). (c) E_P = -ke^2/r = -4.34\times10^{-18}\,\mathrm{J} = -27.1\,\mathrm{eV} (A1); total = 13.6 - 27.1 = -13.6\,\mathrm{eV}, equal to the ground-state energy -13.6/1^2\,\mathrm{eV} (A1). (d) mvr = 9.11\times10^{-31}\times2.18\times10^{6}\times5.3\times10^{-11} = 1.05\times10^{-34}\,\mathrm{J\,s}, and h/2\pi = 1.06\times10^{-34}\,\mathrm{J\,s}: equal within the precision of the data (A1).

(a) E_4 = -13.6/4^2 = -13.6/16 = -0.85\,\mathrm{eV} (A1). (b) E_2 = -13.6/4 = -3.40\,\mathrm{eV}; gap = -0.85 - (-3.40) = 2.55\,\mathrm{eV} (M1) = 2.55\times1.60\times10^{-19} = 4.08\times10^{-19}\,\mathrm{J} (A1); f = E/h = 4.08\times10^{-19}/6.63\times10^{-34} = 6.15\times10^{14}\,\mathrm{Hz} (A1). (c) 1\to2: 13.6(1 - \tfrac14) = 10.2\,\mathrm{eV}; 1\to3: 13.6(1 - \tfrac19) = 12.1\,\mathrm{eV} (M1). Ratio = \tfrac34\big/\tfrac89 = 27/32 = 0.84 (A1). Using n values directly (2/3) is the usual slip.

(a) E_K = \tfrac12mv^2 = 0.5\times6.64\times10^{-27}\times(1.52\times10^{7})^2 = 7.67\times10^{-13}\,\mathrm{J} (M1); \div\,1.60\times10^{-13}\,\mathrm{J\,MeV^{-1}} = 4.79\,\mathrm{MeV} (A1). (b) When the alpha particle stops for an instant, its starting kinetic energy equals the electric potential energy of the pair: E_K = k(2e)(79e)/d (M1). d = 2\times79\times8.99\times10^{9}\times(1.60\times10^{-19})^2/7.67\times10^{-13} (A1) = 4.74\times10^{-14}\,\mathrm{m} (A1). (c) R = R_0A^{1/3} = 1.20\times10^{-15}\times197^{1/3} = 6.98\times10^{-15}\,\mathrm{m} (A1). d/R \approx 6.8: the alpha particle turns back well outside the nucleus, so only the electric force acts on it (A1).

(a) V = \tfrac43\pi R^3 = \tfrac43\pi R_0^3A (M1). \rho = \dfrac{A\times1.67\times10^{-27}}{\tfrac43\pi R_0^3A} = \dfrac{3\times1.67\times10^{-27}}{4\pi R_0^3}: A cancels (A1). (b) \rho = 3\times1.67\times10^{-27}/(4\pi\times(1.20\times10^{-15})^3) (M1) = 2.3\times10^{17}\,\mathrm{kg\,m^{-3}} (A1). (c) V = \tfrac43\pi(1.2\times10^{4})^3 = 7.24\times10^{12}\,\mathrm{m^3} (M1); M = \rho V = 2.31\times10^{17}\times7.24\times10^{12} = 1.7\times10^{30}\,\mathrm{kg}, close to the mass of the Sun (A1).

(a) E = hc/\lambda = 1.989\times10^{-25}/(248.6\times10^{-9}) = 8.00\times10^{-19}\,\mathrm{J} = 5.00\,\mathrm{eV} (M1); in the same way 414.4\,\mathrm{nm} gives 3.00\,\mathrm{eV} and 621.6\,\mathrm{nm} gives 2.00\,\mathrm{eV} (A1). (b) With three levels the two small gaps must add up to the large one: 2.0 + 3.0 = 5.0\,\mathrm{eV} (A1). (c) Emission tells us the sizes of the two steps but not their order: the middle level could be 2.0\,\mathrm{eV} or 3.0\,\mathrm{eV} above the lowest (A1). (d) A cool gas absorbs only from the lowest level, so the two dark lines are the jumps from -7.4\,\mathrm{eV} (M1). 621.6\,\mathrm{nm} (2.0\,\mathrm{eV}): middle level = -7.4 + 2.0 = -5.4\,\mathrm{eV} (A1). 248.6\,\mathrm{nm} (5.0\,\mathrm{eV}): top level = -7.4 + 5.0 = -2.4\,\mathrm{eV} (A1).

(a) With the positive charge spread over a sphere 10^{-10}\,\mathrm{m} across, the electric field anywhere in or near the atom is weak (M1), so every alpha particle passes through with at most a small deflection (A1). (b) A small fraction were deflected through very large angles, some through more than 90^\circ (M1). Reversing a fast alpha particle needs a very strong repulsion, which requires the positive charge to be concentrated in a region far smaller than the atom (A1). (c) The fraction decreases (M1). The force \propto Z at a given distance, so an alpha particle must pass much closer to an aluminium nucleus to be turned through a large angle, and fewer alpha particles pass that close (A1).

(a) [h] = \mathrm{J\,s} = \mathrm{kg\,m^2\,s^{-1}} and [mvr] = \mathrm{kg\times m\,s^{-1}\times m} = \mathrm{kg\,m^2\,s^{-1}} (A1). (b) n = 2\pi L/h = 2\pi\times4.22\times10^{-34}/6.63\times10^{-34} = 4.00, so n = 4 (A1); E_4 = -13.6/16 = -0.85\,\mathrm{eV} (A1). (c) Levels n = 1 to 4 give 4\times3/2 = 6 frequencies (A1). (d) Largest gap 4\to1: 13.6(1 - \tfrac1{16}) = 12.75\,\mathrm{eV} = 2.04\times10^{-18}\,\mathrm{J} (M1); f = 2.04\times10^{-18}/6.63\times10^{-34} = 3.08\times10^{15}\,\mathrm{Hz} (A1). (e) Visible lines end on n = 2 (M1): 4\to2, 2.55\,\mathrm{eV}, \lambda = 488\,\mathrm{nm}; 3\to2, 1.89\,\mathrm{eV}, \lambda = 658\,\mathrm{nm}. Jumps to n = 1 are ultraviolet and 4\to3 (1880\,\mathrm{nm}) is infrared (A1).

(a) The only force is the electric (Coulomb) repulsion between the alpha particle and the nucleus, both treated as point charges (A1). (b) A more energetic alpha particle gets closer before it turns back (M1). Above about 8.0\,\mathrm{MeV} it comes within the short range of the strong nuclear force, which the Rutherford model leaves out, so the scattering changes (A1). (c) d = k(2e)(13e)/E_K = 8.99\times10^{9}\times26\times(1.60\times10^{-19})^2/(8.0\times1.60\times10^{-13}) (M1) = 4.7\times10^{-15}\,\mathrm{m} (A1). (d) R = 1.20\times10^{-15}\times27^{1/3} = 3.6\times10^{-15}\,\mathrm{m}, about three quarters of d (A1). The alpha particle is not a point: it has a radius of its own (about 1.9\times10^{-15}\,\mathrm{m}), and the strong force reaches a little beyond each surface, so the onset comes before the centres are R apart (A1).

(a) Cubing the radius law gives V = \tfrac43\pi R^3 = \tfrac43\pi R_0^3A, a constant times A with no constant term (A1). (b) V = \tfrac43\pi(7.32\times10^{-15})^3 = 1.64\times10^{-42}\,\mathrm{m^3} (A1). (c) Gradient = \tfrac43\pi R_0^3, the volume per nucleon (M1); R_0 = (3\times7.61\times10^{-45}/4\pi)^{1/3} = 1.22\times10^{-15}\,\mathrm{m} (A1). (d) Mass per nucleon divided by volume per nucleon (M1): \rho = 1.67\times10^{-27}/7.61\times10^{-45} = 2.19\times10^{17}\,\mathrm{kg\,m^{-3}} (A1).

(a) Light from the hot surface passes through cooler gas in the star's atmosphere, whose atoms absorb only photons with energies equal to their level gaps (M1); the absorbed energy is re-emitted in all directions, so those wavelengths are weak in the light that reaches us (A1). (b) E.1: E = hc/\lambda = 1.989\times10^{-25}/486.1\times10^{-9} = 4.09\times10^{-19}\,\mathrm{J} (M1) = 2.56\,\mathrm{eV} (A1). (c) C.5: v = c\,\Delta\lambda/\lambda = 3.00\times10^{8}\times0.8/486.1 = 4.9\times10^{5}\,\mathrm{m\,s^{-1}} (A1); the wavelength is longer (redshift), so the star is moving away (A1). (d) B.1: T = 2.9\times10^{-3}/\lambda_{\max} = 2.9\times10^{-3}/414\times10^{-9} (M1) = 7.0\times10^{3}\,\mathrm{K} (A1).

(a) E.1: gap = -3.03 - (-5.13) = 2.10\,\mathrm{eV} = 3.36\times10^{-19}\,\mathrm{J} (M1); \lambda = hc/E = 1.989\times10^{-25}/3.36\times10^{-19} = 592\,\mathrm{nm} (A1). (b) B.5: P = IV = 0.35\times230 = 80.5\,\mathrm{W} (A1). (c) Light power = 0.25\times80.5 = 20.1\,\mathrm{W} (M1); photons per second = 20.1/3.36\times10^{-19} = 6.0\times10^{19}\,\mathrm{s^{-1}} (A1). (d) The atom's energy levels are discrete, so every photon from this transition carries the same energy (M1), and E = hc/\lambda then gives one wavelength, which appears as one bright line (A1).

(a) E.3: the alpha particle is ^{4}_{2}\alpha (A1); lead keeps A = 210 - 4 = 206 and Z = 84 - 2 = 82: ^{210}_{84}\mathrm{Po}\to{}^{206}_{82}\mathrm{Pb} + {}^{4}_{2}\alpha (A1). (b) A.3: E_K = 5.30\times1.60\times10^{-13} = 8.48\times10^{-13}\,\mathrm{J} (M1); v = \sqrt{2E_K/m} = \sqrt{2\times8.48\times10^{-13}/6.64\times10^{-27}} = 1.60\times10^{7}\,\mathrm{m\,s^{-1}} (A1). (c) D.2 and E.1: E_K = k(2e)(79e)/d (M1); d = 8.99\times10^{9}\times158\times(1.60\times10^{-19})^2/8.48\times10^{-13} (A1) = 4.29\times10^{-14}\,\mathrm{m} (A1). (d) Alpha particles lose their energy in collisions with air molecules within a few centimetres, so without a vacuum they would not reach the foil with a known energy (A1).

(a) E.1: E_2 - E_1 = -13.6/4 - (-13.6) = 10.2\,\mathrm{eV} (A1). (b) E = 10.2\times1.60\times10^{-19} = 1.63\times10^{-18}\,\mathrm{J} (M1); \lambda = hc/E = 1.989\times10^{-25}/1.63\times10^{-18} = 1.22\times10^{-7}\,\mathrm{m} (A1). (c) E.2: E_{\max} = hf - \Phi = 10.2 - 4.3 = 5.9\,\mathrm{eV} (M1) = 9.4\times10^{-19}\,\mathrm{J} (A1). (d) p = \sqrt{2mE_{\max}} = \sqrt{2\times9.11\times10^{-31}\times9.44\times10^{-19}} = 1.31\times10^{-24}\,\mathrm{kg\,m\,s^{-1}} (M1); \lambda = h/p = 5.1\times10^{-10}\,\mathrm{m} (A1). (e) An atom is about 10^{-10}\,\mathrm{m} across, so the wavelength is about five atomic diameters: the same order of magnitude, which is why such electrons diffract from crystals (A1).