OCR GCSE Physics A: Ionisation, half-life and penetrating power (J249)

This guide follows the OCR GCSE (9-1) Gateway Science Physics A specification (J249), Version 5.0 (July 2026), for first assessment in 2018. It covers statements P6.1h to P6.1l in topic P6.1, Radioactive emissions. Statements P6.1h, P6.1i, P6.1j and P6.1l are assessed in both Foundation and Higher Tier papers. Statement P6.1k is printed in bold in the specification, so it is Higher Tier only. Nuclear structure, isotopes and decay equations (P6.1a to P6.1g) are taught in a separate unit and are not repeated here.

Related pages: revision notes, practice questions, the OCR GCSE Physics hub, the printable checklist and the free diagnostics.

What this unit covers

Statement	What you must be able to do
P6.1h	Recall that electrons sit at different distances from the nucleus, that absorbing or emitting electromagnetic radiation can change this arrangement, and that losing outer electrons makes an ion
P6.1i	Recall that changes in atoms and nuclei generate and absorb radiation over a wide frequency range
P6.1j	Explain half-life and link it to the random nature of decay
P6.1k (Higher Tier only)	Calculate the net decline, as a ratio, after a whole number of half-lives, including from half-life graphs
P6.1l	Recall how alpha particles, beta particles and gamma rays differ in penetration

Electron energy levels and ionisation (P6.1h)

An atom has a small, positive nucleus. The electrons are arranged at different distances from it. Each arrangement corresponds to a different energy level. Electrons closer to the nucleus are at lower energy levels than electrons further out.

Two things can happen to an electron when radiation interacts with the atom.

• Absorption. If an electron absorbs energy from electromagnetic radiation, it can rise to a higher energy level. The electron is then described as excited. The specification notes that inner electrons can be excited in this way.
• Emission. When the electron loses that energy, it falls back to a lower level and the energy leaves the atom as electromagnetic radiation.

A larger drop between levels releases radiation of higher frequency. A small drop releases radiation of lower frequency.

Ionisation. If an outer electron is removed from an atom altogether, the process is called ionisation. The atom has lost negative charge, so it is left with an overall positive charge. A charged atom is an ion.

Keep two ideas apart. Excitation moves an electron to a higher level and the electron stays in the atom. Ionisation removes the electron from the atom.

Practical link. The specification suggests demonstrating fluorescence with a black light lamp and tonic water. The black light supplies ultraviolet radiation, which electrons in the liquid absorb. When they lose the energy, they emit visible light, and the tonic water glows.

Worked example 1: describing a change in an atom

An atom absorbs a quantity of radiation and one of its electrons moves from a lower to a higher energy level. Later, the atom emits radiation. Describe what has happened to the electron between the two events.

1. The electron absorbed energy from the radiation and became excited, moving to a higher level.
2. The electron then lost that energy and returned to a lower level.
3. The energy lost was emitted as electromagnetic radiation.

Nothing in this example is ionisation, because the electron never left the atom.

Radiation from atoms and nuclei (P6.1i)

Changes in atoms and in nuclei can both produce and absorb radiation. The radiation can come from any part of the electromagnetic spectrum, including gamma rays.

Source of the change	Type of change	Radiation you may meet
Outer electrons	Moving between levels	Visible light, ultraviolet, infrared
Inner electrons	Moving between levels	Higher-frequency radiation such as X-rays
Nucleus	Rearranging after unstable decay	Gamma rays

The key point is the wide range. Do not describe gamma rays as separate from the electromagnetic spectrum. They are the highest-frequency part of it, and they come from the nucleus rather than from electron movements.

Half-life and randomness (P6.1j)

Some nuclei are unstable and decay. The decay of any one nucleus is random. You cannot predict when a particular nucleus will decay, and you cannot make it decay sooner or later by heating, cooling or squeezing the sample.

Random does not mean unpredictable overall. A sample contains an enormous number of nuclei, so the pattern across the whole sample is steady and can be described by the half-life.

Half-life is the time taken for the number of undecayed nuclei in a sample to halve. It is also the time taken for the activity (the count of decays per second) of the source to halve.

Two points follow from randomness:

• Half-life is a statement about large numbers. It says what happens to half of a big sample, not which half.
• Each half-life halves whatever is left. It does not remove a fixed amount.

Modelling decay with dice or coins

The specification suggests using dice to model random decay. Here is a coin version. Start with 64 coins. Throw them all, then remove every coin that lands tails. Each throw stands for one half-life, because each coin has a one-in-two chance of being removed. Over many throws the counts fall roughly 64, 32, 16, 8, 4. The real results will wobble around those values, and with very few coins left they will wobble a lot. This shows why half-life is reliable for large samples and poor for tiny ones.

Worked example 2: reading a half-life from data

A source has activity 640 Bq at time zero. Its activity is measured every 6 minutes.

Time / min	0	6	12	18	24	30
Activity / Bq	640	453	320	226	160	113

1. Find the time for the activity to fall from 640 Bq to half, 320 Bq. This is 12 minutes.
2. Check with a second pair: 320 Bq to 160 Bq takes from 12 minutes to 24 minutes, so 12 minutes again.
3. The two intervals agree, so the half-life is 12 minutes.

Using several intervals is better than using one.

Net decline after whole half-lives (P6.1k, Higher Tier only)

After one half-life, one half is left. After each further half-life, the remainder halves again. After n half-lives, the amount left is the original divided by 2ⁿ.

Half-lives passed	Fraction remaining	Ratio of original to remaining	Fraction decayed
1	1/2	2 : 1	1/2
2	1/4	4 : 1	3/4
3	1/8	8 : 1	7/8
4	1/16	16 : 1	15/16
5	1/32	32 : 1	31/32

Method.

1. Divide the elapsed time by the half-life to get the number of half-lives, n.
2. Work out 2ⁿ by doubling n times.
3. Divide the starting value by 2ⁿ for the amount left. Give the ratio if asked.

State what your ratio compares. “Remaining to original is 1 : 16” and “original to remaining is 16 : 1” are different statements.

Half-life graphs. On a graph of activity against time the points lie on a smooth curve. It falls steeply at first, then more slowly, and never reaches zero. To read the half-life, pick an activity, find the time when it has fallen to half, then repeat from a different starting activity. The same time interval each time confirms a constant half-life. To find the value after a whole number of half-lives, step along the time axis one half-life at a time, halving each step.

Worked example 3: activity and ratio

A sample has an activity of 1280 Bq and a half-life of 7.5 hours. Find its activity after 30 hours and the net decline as a ratio.

1. Number of half-lives = 30 ÷ 7.5 = 4.
2. 2⁴ = 16.
3. Activity after 30 hours = 1280 ÷ 16 = 80 Bq.
4. The ratio of original activity to final activity is 1280 : 80, which simplifies to 16 : 1.
5. The fraction that has decayed is 15/16.

Worked example 4: counting nuclei

A sample holds 3.2 × 10⁶ undecayed nuclei. The half-life is 20 days. How many undecayed nuclei remain after 100 days, and how many have decayed?

1. Number of half-lives = 100 ÷ 20 = 5, and 2⁵ = 32.
2. Undecayed nuclei = 3.2 × 10⁶ ÷ 32 = 1.0 × 10⁵.
3. Decayed nuclei = 3.2 × 10⁶ − 1.0 × 10⁵ = 3.1 × 10⁶.

Allowing for background radiation

A detector also counts background radiation from its surroundings. Only the part of the reading that comes from the source halves each half-life. Subtract the background first, apply the halving, then add the background back if you want to predict the reading.

Worked example 5: a reading with background

A detector gives 372 counts per minute with a source present. The background is 20 counts per minute. The source has a half-life of 3 hours. What will the detector read 9 hours later?

1. Count rate from the source = 372 − 20 = 352 counts per minute.
2. Number of half-lives = 9 ÷ 3 = 3, so divide by 8: 352 ÷ 8 = 44 counts per minute.
3. Add the background: 44 + 20 = 64 counts per minute.

Dividing 372 by 8 directly would give the wrong answer, because the background does not decay.

Penetration of alpha, beta and gamma (P6.1l)

The three emissions differ in how far they get through matter.

Radiation	Stopped by	Typical range in air
Alpha particles	A sheet of paper or a few centimetres of air	A few centimetres
Beta particles	A few millimetres of aluminium	Around a metre
Gamma rays	Intensity reduced by thick lead or very thick concrete, never completely stopped	Very long

The order of penetration is alpha, then beta, then gamma, from least to most penetrating. Radiation loses energy as it ionises atoms along its path. Strongly ionising radiation uses up its energy over a short distance. Alpha is the most strongly ionising and gamma the least.

Practical link. The specification suggests using a Geiger-Müller tube, sources and aluminium plates of varying thickness to see how the count rate changes.

Worked example 6: identifying radiation from absorber data

A Geiger-Müller tube is placed a fixed distance from a source. The background count rate is 25 counts per minute.

Absorber placed in front of the tube	Count rate / counts per min
None	410
Paper	408
2 mm aluminium	190
6 mm aluminium	28
20 mm lead, with the aluminium removed	26

1. Subtract the background from each reading: 385, 383, 165, 3, 1.
2. Paper barely changes the reading, so alpha particles are not reaching the tube.
3. Aluminium reduces the corrected count rate greatly, to almost nothing at 6 mm. That is typical of beta particles.
4. Lead leaves almost nothing, so there is no significant gamma component.
5. Conclusion: the source emits beta particles only.

Common errors

• Saying that after two half-lives everything has decayed. One quarter remains.
• Applying halving to the background as well as the source.
• Describing ionisation as an electron moving to a higher level. That is excitation.
• Saying an object becomes radioactive because radiation has passed through it. Irradiating an object does not make it radioactive.

Next steps

Condense the unit with the revision notes, then test yourself with the practice questions. Use the free diagnostics to find weak topics.

Official syllabus

OCR, GCSE (9-1) Gateway Science Physics A (J249) specification, Version 5.0 (July 2026), for first assessment in 2018. Statements P6.1h to P6.1l.