2022 H2 Physics Paper 2 Q6 — Half-Life, a Missing Particle and Strontium-90

The model, live

Recoiling nucleusBeta particle / JAntineutrino / K

Part (a)

1 mark

Define the half-life of a radioactive isotope.

The half-life is the time taken for the number of undecayed nuclei of the isotope (or its activity) to fall to half of its initial value.

Part (b)

1 mark

Explain why it is better to use the steep part of an experimental decay curve to find the half-life.

Where the curve is steep, the count rate changes quickly with time, so the time at which a particular value is reached can be read more precisely (and the count rates are larger, so random fluctuations are a smaller fraction of the reading).

Part (c)

2 marks

A stationary nucleus emits a beta particle and recoils; the beta particle leaves at an angle $\theta$ to the line of recoil. Explain why this shows that another particle is emitted.

Approach

Momentum is conserved. Before the decay it is zero, so afterwards the momenta of all the products must add to zero. Two momenta can only add to zero if they are equal and exactly opposite.

The nucleus was stationary, so the total momentum before the decay was zero. The recoiling nucleus and the beta particle do not move in exactly opposite directions, so their momenta cannot add to zero. Another particle (an antineutrino) must be emitted, carrying the momentum needed to make the total zero.

Part (d)(i)

3 marks

Strontium-90 ( ${}^{90}_{38}\text{Sr}$ ) emits a beta particle to become yttrium. Complete the decay equation, including all the decay products.

{}^{90}_{38}\text{Sr} \;\rightarrow\; {}^{90}_{39}\text{Y} \;+\; {}^{0}_{-1}\beta \;+\; \bar{\nu}

${}^{90}_{38}\text{Sr} \rightarrow {}^{90}_{39}\text{Y} + {}^{0}_{-1}\beta + \bar\nu$ (one mark each for the yttrium nuclide, the beta particle and the antineutrino).

To check your answer:

Nucleon numbers: $90 = 90 + 0$ . Proton (charge) numbers: $38 = 39 + (-1)$ . Both balance.

Part (d)(ii)

2 marks

The graph shows the activity $A$ of a sample of strontium-90 against the number $N$ of its nuclei. Find the half-life.

Approach

$A = \lambda N$ , so the graph is a straight line through the origin with gradient $\lambda$ . Read the gradient from a point far along the line, then use $t_{1/2} = \dfrac{\ln 2}{\lambda}$ .

\lambda = \frac{5.25\times10^{9}}{7.0\times10^{18}} = 7.5\times10^{-10}\,\text{s}^{-1}

t_{1/2} = \frac{\ln 2}{\lambda} = \frac{0.693}{7.5\times10^{-10}} = 9.2\times10^{8}\,\text{s}

half-life $= 9.2\times10^{8}\,\text{s}$

Note

That is about 29 years, the accepted half-life of strontium-90.

Part (e)

3 marks

An unstable isotope J decays to a stable isotope K. For an initially pure sample of J, find the ratio (atoms of J)/(atoms of K) after 3.5 half-lives.

Approach

The fraction of J left after $n$ half-lives is $\left(\tfrac12\right)^n$ , and $n$ need not be a whole number. Every J atom that has decayed is now a K atom, so K is the rest.

\frac{N_J}{N_0} = \left(\tfrac12\right)^{3.5} = 0.0884 \qquad \frac{N_K}{N_0} = 1 - 0.0884 = 0.912

\frac{N_J}{N_K} = \frac{0.0884}{0.912} = 0.097

ratio $= 0.097$

Watch

The question asks for J to K, not J to the original amount. $0.088$ is the fraction remaining; the ratio is $0.097$ .

What this question was really testing

Half-life, defined precisely.
Momentum conservation as evidence. Two products that are not back-to-back need a third.
Balancing a decay equation. Nucleon and proton numbers, plus the antineutrino.
$A = \lambda N$ . The gradient of an A–N graph is the decay constant.
Fractional half-lives. $\left(\tfrac12\right)^{3.5}$ , and parent against daughter rather than against the start.

A recoil that doesn't line up, and a graph whose gradient is the decay constant

Part (a)

Part (b)

Part (c)

Part (d)(i)

Part (d)(ii)

Part (e)

What this question was really testing

This is how we mark, and how we teach.