GCSE Physics (9-1)

9.2 Atoms and nuclear radiation — Practise with help

Work each problem first. Open a hint only if you are stuck, and the full working only to check.

Warm-up: remember

Answer these from memory. They use ideas from Section 9.1 and earlier work on powers and decimals.

  1. How many protons are in an atom with atomic number 20?
  2. Work out 242^4.
  3. Write 18\tfrac{1}{8} as a decimal.
Answers
  1. 20 protons.
  2. 24=162^4 = 16.
  3. 0.1250.125.

If question 1 went wrong, look again at the atomic number in Section 9.1. If question 2 went wrong, remember that a power tells you how many copies to multiply. If question 3 went wrong, divide 1 by 8.

Practise with help

1. Define activity, and state the unit in which it is measured.

Hint

Activity is a rate: it counts decays in each second.

Next step

Say what is decaying, and how often. Then name the unit and what one unit means.

Full working

Activity is the rate at which the unstable nuclei in a source decay. It is measured in becquerels (Bq). One becquerel is one decay each second.

2. Describe an alpha particle, a beta particle and a gamma ray, and say what stops each one.

Hint

One of the three has no mass and no charge at all.

Next step

Say what each particle is made of, then match it to the material that stops it: paper, aluminium or thick lead.

Full working

An alpha particle is two protons and two neutrons, the same as a helium nucleus. It is stopped by paper or skin. A beta particle is a fast electron ejected from the nucleus. It is stopped by a few millimetres of aluminium. A gamma ray is electromagnetic radiation from the nucleus, with no mass and no charge. It is reduced by thick lead or concrete.

3. A nucleus of radon-222, 86222Rn^{222}_{86}\text{Rn}, emits an alpha particle. Write the balanced nuclear equation.

Hint

An alpha particle is 24α^{4}_{2}\alpha, so the mass number falls by 4 and the atomic number falls by 2.

Next step

Work out the new mass number 222−4222 - 4 and the new atomic number 86−286 - 2.

Full working 86222Rn→84218Po+24α^{222}_{86}\text{Rn} \rightarrow {}^{218}_{84}\text{Po} + {}^{4}_{2}\alpha

The mass number is 222−4=218222 - 4 = 218, and the atomic number is 86−2=8486 - 2 = 84.

4. A nucleus of carbon-14, 614C^{14}_{6}\text{C}, emits a beta particle. Write the balanced nuclear equation.

Hint

A beta particle is −10β^{0}_{-1}\beta, so the mass number does not change.

Next step

Keep the mass number at 14, and work out the new atomic number 6+16 + 1.

Full working 614C→714N+−10β^{14}_{6}\text{C} \rightarrow {}^{14}_{7}\text{N} + {}^{0}_{-1}\beta

The mass number stays at 14, and the atomic number goes up by 1 to give 7.

5. A source has a count rate of 320 counts per second. After 12 minutes the count rate is 40 counts per second. Work out the half-life.

Hint

Count how many times the count rate must halve to get from 320 to 40.

Next step

Halve step by step: 320, then 160, then 80, then 40. Count the steps.

Full working

320→160→80→40320 \rightarrow 160 \rightarrow 80 \rightarrow 40 is three halvings, so three half-lives have passed in 12 minutes. The half-life is 123=4.0\dfrac{12}{3} = 4.0 minutes.

6. Explain why the count rate from a source is only roughly halved after one half-life.

Hint

Think about whether you can say which nucleus decays next.

Next step

Say what kind of process decay is, then say what the half-life describes.

Full working

Radioactive decay is random. You cannot predict which nucleus will decay next. The half-life is the average time for half of the nuclei in a large sample to decay, so a real measurement only follows this average roughly.

7. After four half-lives, calculate the net decline in the emission as a ratio.

Hint

Work out the fraction that remains after four halvings, then subtract it from 1.

Next step

Write the fraction remaining as (12)4\left(\dfrac{1}{2}\right)^4, then work out 1−1161 - \dfrac{1}{16}.

Full working

The fraction remaining is (12)4=116\left(\dfrac{1}{2}\right)^4 = \dfrac{1}{16}. The net decline is

1−116=15161 - \frac{1}{16} = \frac{15}{16}

so the net decline is a ratio of 15:1615:16 of the original emission.

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