Physics › Practise › Questions › Nuclear radius and density
Questions on nuclear radius and density
The distance of closest approach of an alpha particle gives an upper limit for nuclear radius, and electron diffraction gives the radius itself. Both point to R = R₀A^(1/3), with R₀ about 1.2 fm from the gradient of R against A^(1/3). Nuclear volume is then proportional to nucleon number, so all nuclei have about the same density.
18 original questions · 49 marks · the nuclear radius and density notes · Nuclear physics
To print: the questions (PDF, 57 KB) · the worked answers (PDF, 76 KB)
This topic is tagged for AQA, Edexcel, OCR A.
Every question here is written for this library rather than taken from a past paper. Write your answer out before opening a mark scheme: the schemes award marks point by point, and the marks are easier to see when you have something of your own to compare against.
State the equation relating nuclear radius to nucleon number, and give the approximate value of the constant R0.
Mark scheme
R = R0A1/3, where A is the nucleon number (1); R0 ≈ 1.2 × 10−15 m (1.2 fm) (1).Calculate the radius of a nucleus with nucleon number A = 64.
R0 = 1.2 × 10−15 mMark scheme
R = R0A1/3 = 1.2 × 10−15 × 641/3 = 1.2 × 10−15 × 4 (1)
R = 4.8 × 10−15 m (1)State what the relationship R = R0A1/3 implies about the density of nuclear matter.
Mark scheme
Volume ∝ R3 ∝ A, and mass ∝ A (1); so the density (mass/volume) is approximately the same for all nuclei (1).State typical values for the radius of an atom and for the radius of a nucleus.
Mark scheme
Atomic radius: about 1 × 10−10 m (1). Nuclear radius: a few femtometres, of order 10−15 m to 10−14 m (1).A student has measured nuclear radii R for nuclides covering a wide range of nucleon numbers A. State the graph the student should plot to obtain a straight line through the origin, and state what the gradient of that line represents.
Mark scheme
Plot R against A1/3 (1); the gradient of the line is the constant R0, about 1.2 fm (1).Calculate the radius of an aluminium nucleus, for which A = 27.
R0 = 1.2 × 10−15 mMark scheme
R = R0A1/3 = 1.2 × 10−15 × 271/3 = 1.2 × 10−15 × 3 (1)
R = 3.6 × 10−15 m (1)Taking the mass of a nucleon as 1.66 × 10−27 kg and the volume per nucleon as that of a sphere of radius R0, estimate the density of nuclear matter.
R0 = 1.2 × 10−15 mMark scheme
ρ = u/((4/3)πR03) (1)
= (1.66 × 10−27)/((4/3)π × (1.2 × 10−15)3) (1)
ρ = 2.3 × 1017 kg m−3 (1)Calculate the ratio of the radius of a nucleus with A = 125 to the radius of a nucleus with A = 8.
Mark scheme
R125/R8 = (125/8)1/3 (1)
ratio = 2.5 (1)Explain how the near-constant density of nuclei supports the idea that nucleons are closely packed.
Mark scheme
Constant density means each nucleon occupies about the same volume in every nucleus (1); this is what is expected if nucleons are in contact, packed like touching spheres (1); if nucleons were spread out, larger nuclei could have lower densities, which is not observed (1).Electron diffraction shows that a nuclide has a nuclear radius of 6.0 × 10−15 m. Determine the nucleon number of the nuclide.
R0 = 1.2 × 10−15 mMark scheme
A = (R/R0)3 = (6.0/1.2)3 (1)
A = 125 (1)The radius of a gold atom is 1.35 × 10−10 m and the radius of a gold nucleus is 7.0 × 10−15 m. Calculate the fraction of the volume of the atom occupied by the nucleus, and comment on your answer.
Mark scheme
fraction = (Rnucleus/Ratom)3 = (7.0 × 10−15/(1.35 × 10−10))3 (1)
fraction = 1.4 × 10−13 (1)
The nucleus occupies a vanishingly small fraction of the atom's volume: the atom is almost entirely empty space, as Rutherford scattering showed (1)Explain why high-energy electrons are a suitable probe for measuring the radius of a nucleus.
Mark scheme
At high energies the de Broglie wavelength of an electron is comparable to the diameter of a nucleus, so the nucleus produces a measurable diffraction pattern (1). Electrons are leptons, so they do not experience the strong nuclear force and probe the charge distribution alone (1).An alpha particle with kinetic energy 5.0 MeV is fired head-on at a gold nucleus (Z = 79). Estimate the distance of closest approach.
k = 8.99 × 109 N m2 C−2, e = 1.60 × 10−19 CMark scheme
At closest approach all the kinetic energy has become electrostatic potential energy: Ek = k(2e)(79e)/r (1)
Ek = 5.0 × 106 × 1.60 × 10−19 = 8.0 × 10−13 J (1)
r = k × 2 × 79 × e2/Ek (1)
r = 4.5 × 10−14 m (1)Calculate the density of a carbon-12 nucleus (A = 12), taking the nucleon mass as 1.66 × 10−27 kg and R0 = 1.2 × 10−15 m. Compare your value with the density of nuclear matter in general, 2.3 × 1017 kg m−3.
Mark scheme
R = 1.2 × 10−15 × 121/3 = 2.75 × 10−15 m (1)
mass = 12 × 1.66 × 10−27 = 1.99 × 10−26 kg (1)
ρ = m/((4/3)πR3) = 2.3 × 1017 kg m−3 (1)
This is the same as the density of nuclear matter in general, as expected for constant nuclear density (1).Describe how electron diffraction can be used to determine the radius of a nucleus.
Mark scheme
High-energy electrons, with a de Broglie wavelength comparable to the nuclear diameter, are fired at a thin sample (1). The electrons diffract around the nuclei, giving an intensity pattern with minima (1). The angle of the first minimum depends on the nuclear diameter, so measuring it allows the radius to be calculated (1).Determine the minimum initial kinetic energy, in MeV, that an alpha particle needs in order to approach within 4.0 × 10−14 m of the centre of a gold nucleus (Z = 79) in a head-on collision.
k = 8.99 × 109 N m2 C−2, e = 1.60 × 10−19 C, 1 MeV = 1.60 × 10−13 JMark scheme
At closest approach all the kinetic energy has become electrostatic potential energy: Ek = k(2e)(79e)/r (1)
Ek = 8.99 × 109 × 2 × 79 × (1.60 × 10−19)2/(4.0 × 10−14) (1)
Ek = 9.1 × 10−13 J (1)
Ek = 9.1 × 10−13/(1.60 × 10−13) = 5.7 MeV (1)For one nuclide, the closest-approach method with 5 MeV alpha particles gives a nuclear radius estimate of 4.5 × 10−14 m, while electron diffraction gives 7.2 × 10−15 m. Explain why the two values differ, and deduce which is the better estimate of the nuclear radius.
Mark scheme
The alpha particle stops where all its kinetic energy has become electrostatic potential energy, which is before it reaches the nuclear surface (1); closest approach therefore gives only an upper limit, and the value depends on the alpha energy used rather than on the nucleus alone (1). Electron diffraction measures the nucleus itself: the angle of the first minimum of the diffraction pattern is set by the nuclear diameter (1). The electron diffraction value of 7.2 × 10−15 m is therefore the better estimate (1).A neutron star can be modelled as a sphere with the density of nuclear matter, 2.3 × 1017 kg m−3. Estimate the radius of a neutron star of mass 4.0 × 1030 kg.
Mark scheme
V = m/ρ = (4.0 × 1030)/(2.3 × 1017) = 1.7 × 1013 m3 (1)
V = (4/3)πR3, so R = (3V/4π)1/3 (1)
R = 1.6 × 104 m (about 16 km) (1)
The same practice on paper: the printable workbook for this topic, questions and a worked answer book.
Practise nuclear radius and density one question at a time
The player marks nothing for you. It shows one question, waits, then shows the scheme so you can mark yourself, and brings a question back sooner when it went badly.