Nuclear radius, mass–energy, fission and safety – A-level Physics exam questions with mark scheme

1Hard6 marks

The radius $R$ of a nucleus with nucleon number $A$ is given by $R = r_0A^{1/3}$, where $r_0$ = 1.2 fm.

(a) Calculate the radius of a gold-197 nucleus. [1]

(b) Show that the density of nuclear matter does not depend on $A$, and calculate its value. Take the mass of a nucleon as 1.661 × 10$^{-27}$ kg. [3]

(c) In a scattering experiment, alpha particles of kinetic energy 5.0 MeV are directed at gold nuclei (proton number 79). Calculate the distance of closest approach of an alpha particle travelling head-on towards a gold nucleus. [2]

(a) $R = 1.2 \times 197^{1/3}$ = 7.0 fm (6.98 × 10$^{-15}$ m)(1)
(b) $\rho = \frac{Am}{\frac{4}{3}\pi r_0^3 A}$ — volume ∝ $R^3$ ∝ $A$ and mass ∝ $A$(1)
(b) so $A$ cancels: $\rho = \frac{3m}{4\pi r_0^3}$, independent of $A$(1)
(b) $\rho$ = 2.3 × 10$^{17}$ kg m$^{-3}$(1)
(c) KE = electric PE at closest approach: $5.0\times1.60\times10^{-13} = \frac{(79e)(2e)}{4\pi\varepsilon_0 r}$(1)
(c) $r$ = 4.5 × 10$^{-14}$ m (4.55)(1)
Answer: (a) 7.0 fm (b) 2.3 × 10$^{17}$ kg m$^{-3}$ (c) 4.5 × 10$^{-14}$ m
Examiner tip: Convert MeV to joules (× 1.60 × 10⁻¹³) and use charge 2e for the alpha particle.

2Standard5 marks

(a) Define the binding energy of a nucleus. [1]

(b) Calculate the binding energy, in MeV, of a helium-4 nucleus.

mass of helium-4 nucleus = 4.00151 u; mass of proton = 1.00728 u; mass of neutron = 1.00867 u; 1 u is equivalent to 931.5 MeV. [3]

(a) the energy needed to separate a nucleus into its individual (separate) protons and neutrons (accept: energy released when the nucleus forms from its nucleons)(1)
(b) total mass of nucleons $= 2(1.00728) + 2(1.00867)$ = 4.03190 u(1)
(b) mass defect $= 4.03190 - 4.00151$ = 0.03039 u(1)
(b) binding energy $= 0.03039 \times 931.5$ = 28.3 MeV(1)
(c) $28.3 / 4$ = 7.1 MeV per nucleon (7.08)(1)
Answer: (b) 28.3 MeV (c) 7.08 MeV
Examiner tip: Keep all the decimal places of the masses until the subtraction — rounding early wipes out the mass defect.

3Hard6 marks

The graph shows how the binding energy per nucleon varies with nucleon number $A$.

(a) Use the graph to explain why energy is released both when light nuclei fuse and when very heavy nuclei undergo fission. [3]

(b) One possible fission reaction is

$$^{235}_{92}\text{U} +\ ^{1}_{0}\text{n} \rightarrow\ ^{141}_{56}\text{Ba} +\ ^{92}_{36}\text{Kr} + 3\,^{1}_{0}\text{n}$$

The binding energies per nucleon are: U-235 7.59 MeV, Ba-141 8.33 MeV, Kr-92 8.51 MeV. Calculate the energy released in this fission. [3]

(a) energy is released when the products have a greater (total) binding energy than the original nucleus / products have higher binding energy per nucleon(1)
(a) fusion: light nuclei (A below about 56) combine to make nuclei further up the curve, nearer the peak, with higher binding energy per nucleon(1)
(a) fission: heavy nuclei split into medium nuclei closer to the peak (iron/nickel region), which have higher binding energy per nucleon(1)
(b) binding energy of products $= (141\times8.33) + (92\times8.51) = 1174.5 + 782.9$ = 1957.5 MeV(1)
(b) binding energy of U-235 $= 235\times7.59$ = 1783.7 MeV(1)
(b) energy released = 1957.5 − 1783.7 = 174 MeV (allow 170–174 MeV; ignore free neutrons, which have no binding energy)(1)
Answer: (b) 174 MeV
Examiner tip: Multiply binding energy per nucleon by the nucleon number of each nucleus before subtracting — free neutrons have zero binding energy.

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