These questions revise the lessons in nuclear physics. Use c = 3.0 × 10⁸ m s⁻¹ where needed.
Try each question on paper first.
Questions
Question 1
A source’s net count falls from 400 to 40 per minute when a sheet of paper is placed in front of it, and stays at about 40 when aluminium is added. Which radiation is present and which is absent?
Answer
Paper removes 360 of the 400 counts, so alpha is present.
Aluminium changes nothing further, so beta is absent. The remaining 40 counts are probably gamma, which only thick lead would weaken.
Question 2
The net activity of a sample is 800 Bq at time zero and 100 Bq after 18 hours. Find the half-life.
Answer
800 to 100 is three halvings (800, 400, 200, 100). Half-life = 18 ÷ 3 = 6.0 hours.
Question 3
The measured count rate of a source is 330 per minute and the background is 30 per minute. After one half-life the measured rate is 180 per minute. Show that the time is one half-life.
Answer
Net at the start = 330 − 30 = 300.
Net later = 180 − 30 = 150. The net count is exactly half, so the time is one half-life.
The raw values (330 and 180) are not in the ratio 2 : 1, which is why the background must be subtracted first.
Question 4
Complete the equation for alpha decay: ²²⁶₈₈Ra → ?₈₆Rn + ⁴₂He.
Answer
Nucleon number: 226 = ? + 4, so ? = 222.
Proton number: 88 = 86 + 2, which balances. The radon nucleus is ²²²₈₆Rn.
Question 5
Carbon-14 (¹⁴₆C) decays by beta emission. Write the equation and find the nucleon and proton numbers of the new nucleus.
Answer
¹⁴₆C → ¹⁴₇N + ⁰₋₁e.
Nucleon number: 14 = 14 + 0.
Proton number: 6 = 7 + (−1). The new nucleus has nucleon number 14 and proton number 7.
Question 6
In one fission, ²³⁵₉₂U absorbs a neutron and splits into ¹⁴¹₅₆Ba, ⁹²₃₆Kr and some neutrons. How many neutrons are released?
Answer
Nucleon number: 235 + 1 = 236 on the left.
On the right, 141 + 92 = 233, so 236 − 233 = 3 neutrons. Check protons: 92 = 56 + 36.
3 neutrons are released.
Question 7
A fusion reaction converts a mass of 3.0 × 10⁻²⁹ kg into energy. Find the energy released.
Answer
E = mc² = 3.0 × 10⁻²⁹ × (3.0 × 10⁸)² = 3.0 × 10⁻²⁹ × 9.0 × 10¹⁶ = 2.7 × 10⁻¹² J.
Question 8
A hospital tracer for imaging is chosen to have a half-life of a few hours, not a few years. Give one reason.
Answer
A short half-life means the activity falls to a low level soon after the scan, so the body receives radiation for a shorter time. It also leaves little radioactive material to dispose of afterwards.
If you got these wrong
Question 1 needs distinguishing types of radiation. Questions 2 and 3 need interpreting half-life graphs.
Questions 4 and 5 need balancing simple nuclear equations. Questions 6 and 7 need comparing fission and fusion, and question 8 needs applications and risks.
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