Questions on stable and unstable nuclei

Question 1

Describe the strong nuclear force, including its approximate range and whether it is attractive or repulsive.

Accepted Answer

The strong nuclear force holds nucleons together. It is attractive over separations from about 0.5 fm to about 3 fm (1), and becomes repulsive at very small separations (below about 0.5 fm), which stops the nucleus collapsing (1).

Question 2

State how the mass number A and the proton number Z change when a nucleus emits an alpha particle.

Accepted Answer

A decreases by 4 (1) and Z decreases by 2 (1).

Question 3

State what a beta-minus particle is.

Accepted Answer

A beta-minus particle is a fast-moving electron created in and emitted from the nucleus (1) when a neutron changes into a proton (1).

Question 4

State the two quantities that must balance on each side of any nuclear decay equation.

Accepted Answer

The nucleon (mass) number A (1) and the proton number Z, i.e. the charge (1).

Question 5

State what an alpha particle consists of, and write the alpha particle in nuclide notation.

Accepted Answer

Two protons and two neutrons, i.e. a helium-4 nucleus (1). Notation: ⁴₂He (1).

Question 6

Uranium-238 decays by alpha emission to thorium (Th). Write the balanced nuclear equation, including mass and proton numbers.

Accepted Answer

²³⁸₉₂U → ²³⁴₉₀Th (1) + ⁴₂He (the alpha particle) (1). Mass numbers: 238 = 234 + 4 (1); proton numbers: 92 = 90 + 2 (1).

Question 7

Carbon-14 decays by beta-minus emission to nitrogen (N). Write the balanced nuclear equation, including the antineutrino.

Accepted Answer

¹⁴₆C → ¹⁴₇N (1) + ⁰₋₁e (1) + anti-ν_e (1). Mass numbers: 14 = 14 + 0; charge: 6 = 7 + (−1) (1).

Question 8

Explain why the existence of the (anti)neutrino was proposed to account for observations of beta-minus decay.

Accepted Answer

Beta particles are emitted with a range of kinetic energies up to a maximum, rather than a single fixed value (1). To conserve energy (and momentum) (1), a second, unseen particle must carry away the balance; this was the antineutrino (1).

Question 9

In an unknown decay, the proton number increases by 1 while the mass number stays the same. State the type of decay and explain your reasoning.

Accepted Answer

Beta-minus decay (1). A neutron becomes a proton, so Z rises by 1 (1) while A is unchanged (a proton and neutron have the same mass number), and an electron is emitted (1).

Question 10

In one section of a natural decay series, a bismuth-214 nucleus (Z = 83) decays by beta-minus emission, and the daughter nucleus then decays by alpha emission. Determine the nucleon number and the proton number of the final nucleus.

Accepted Answer

Beta-minus: A is unchanged and Z rises by 1, giving A = 214, Z = 84 (1). Alpha: A falls by 4 and Z falls by 2 (1), giving A = 210 and Z = 82 (1).

Question 11

A freshly produced nuclide contains significantly more neutrons than the stable isotopes of the same element. State the decay mode expected and explain how the decay moves the nuclide towards stability.

Accepted Answer

Beta-minus decay (1). A neutron changes into a proton (1), so the neutron number falls by one while the proton number rises by one, reducing the neutron excess (1).

Question 12

In the beta-minus decay of carbon-14 the total energy released is 0.156 MeV. In one particular decay the emitted electron is measured to carry 0.049 MeV of kinetic energy. Calculate the energy carried away by the antineutrino. Assume that the recoil energy of the nucleus is negligible.

Accepted Answer

The antineutrino carries the difference between the total energy released and the electron's share (1). E = 0.156 − 0.049 (1) = 0.107 MeV (1).

Question 13

Radon-220 (Z = 86) decays by alpha emission to a nuclide X. Determine the nucleon number and the proton number of X, and identify X from the following elements: lead (Z = 82), bismuth (Z = 83), polonium (Z = 84), astatine (Z = 85).

Accepted Answer

A = 220 − 4 = 216 (1). Z = 86 − 2 = 84 (1). Z = 84 is polonium, so X is polonium-216 (1).

Question 14

Radium-226 decays by alpha emission to radon (Rn). Write the balanced nuclear equation and state the number of neutrons in the radon nucleus produced.

Accepted Answer

²²⁶₈₈Ra → ²²²₈₆Rn (1) + ⁴₂He (1). The radon nucleus has 222 − 86 (1) = 136 neutrons (1).

Question 15

Describe how the strong nuclear force between two nucleons varies with their separation, referring to the separations at which it is repulsive, attractive and effectively zero.

Accepted Answer

At separations below about 0.5 fm the force is strongly repulsive (1). From about 0.5 fm to about 3 fm it is attractive (1), peaking then falling (1). Beyond about 3 fm it is effectively zero (1), because the force is very short-ranged.

Question 16

Explain why very large nuclei tend to be unstable.

Accepted Answer

The strong force acts only over a very short range, so a nucleon is attracted only to its nearest neighbours (1), whereas the electrostatic repulsion between protons acts across the whole nucleus (1). In large nuclei the long-range repulsion of many protons outweighs the short-range attraction, making them unstable (1).

Question 17

A student measures the kinetic energies of the electrons emitted by a radioactive source and finds that every electron carries the same energy. Deduce whether these electrons can be beta-minus particles.

Accepted Answer

In beta-minus decay the fixed energy released is shared between the electron and the antineutrino (1), so beta-minus electrons show a continuous spread of energies from near zero up to a maximum (1). A single fixed energy is inconsistent with this, so the electrons cannot be beta-minus particles (1).

Question 18

The nuclide used in domestic smoke detectors, americium-241 (Z = 95), decays by alpha emission to neptunium (Np). A data sheet claims that the daughter nucleus produced contains more than 140 neutrons. Deduce whether the claim is correct.

Accepted Answer

Daughter nucleon number A = 241 − 4 = 237 (1). Daughter proton number Z = 95 − 2 = 93 (1). Neutrons = 237 − 93 = 144 (1). 144 > 140, so the claim is correct (1).

Question 19

Compare alpha decay with beta-minus decay. In your answer you should describe what is emitted in each case, state how the nucleon number and the proton number change, and explain the role of the antineutrino in beta-minus decay.

Accepted Answer

Alpha decay: the nucleus emits an alpha particle, two protons and two neutrons bound as a helium-4 nucleus (1). A falls by 4 and Z falls by 2, so a new element two places lower in the periodic table is formed (1). Beta-minus decay: a neutron in the nucleus changes into a proton (1), and a fast electron (the beta particle) is created and emitted together with an electron antineutrino (1). A is unchanged and Z rises by 1 (1). The antineutrino shares the released energy with the electron, which is why beta particles are emitted with a continuous range of energies up to a maximum (1).