IGCSE Physics Unit 5 — Nuclear Physics Numericals (Solutions)

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IGCSE Physics 0625 · Section 5 · Nuclear Physics — Numerical Solutions

Nuclear Physics — Numerical Practice: Solutions

A. Nuclide Notation

A1

A=23, Z=11. Find protons, neutrons, electrons.

1
neutrons = A − Z
23 − 11 = 12
✓Protons = 11, Neutrons = 12, Electrons = 11
A2

A=40, Z=18. Find neutrons.

1
neutrons = A − Z
40 − 18
✓22
A3

6 protons, 8 neutrons. Find A.

1
A = protons + neutrons
6 + 8
✓14 (this is carbon-14)
A4

A=35 and 37, both Z=17. Find neutrons in each.

1
neutrons = A − Z, for each isotope
35−17=18;   37−17=20
✓18 neutrons and 20 neutrons
A5

92 protons, A=238. Find neutrons.

1
neutrons = A − Z
238 − 92
✓146

B. Half-Life Calculations

B1

640 counts/min, half-life 5 days. Find activity after 20 days.

1
Number of half-lives
20 ÷ 5 = 4
2
Halve 4 times
640 → 320 → 160 → 80 → 40
✓40 counts/min
B2

800 → 100 counts/min over 45 min. Find half-life.

1
Count the halvings
800→400→200→100 (3 halvings)
2
half-life = total time / number of halvings
45 ÷ 3
✓15 minutes
B3

Half-life 8 years, 2000 Bq initial. Find activity after 24 years.

1
Number of half-lives
24 ÷ 8 = 3
2
Halve 3 times
2000 → 1000 → 500 → 250
✓250 Bq
B4

Activity drops to 1/8 after 60 min. Find half-life.

1
1/8 = (1/2)³, so 3 half-lives have passed
half-life = 60 ÷ 3
✓20 minutes
B5

50 counts/min after 3 half-lives. Find original activity.

1
Work backward: double 3 times
50 × 2³ = 50 × 8
✓400 counts/min
B6

Why is a very long half-life unsuitable as a medical tracer?

✓A very long half-life means the isotope keeps emitting radiation inside the patient’s body for a very long time after the scan is complete, causing unnecessary prolonged radiation exposure. A shorter half-life ensures the radioactivity decays away soon after the diagnostic information is obtained.

C. Decay Equations and Background Correction

C1

Measured 95, background 15. Find corrected count rate.

1
corrected = measured − background
95 − 15
✓80 counts/min
C2

Corrected 120, measured 138. Find background.

1
background = measured − corrected
138 − 120
✓18 counts/min
C3

Ra-226 (Z=88) → Rn (Z=86) by alpha emission.

1
Write the equation, balancing A and Z
²²⁶₈₈Ra → ²²²₈₆Rn + ⁴₂He
2
Check the balance
A: 226=222+4 ✓   Z: 88=86+2 ✓
✓²²⁶₈₈Ra → ²²²₈₆Rn + ⁴₂He
C4

C-14 (Z=6) → N (Z=7) by beta emission.

1
Write the equation, balancing A and Z
¹⁴₆C → ¹⁴₇N + ⁰₋₁e
2
Check the balance
A: 14=14+0 ✓   Z: 6=7+(−1) ✓
✓¹⁴₆C → ¹⁴₇N + ⁰₋₁e
C5

Why doesn’t nucleon number change in beta decay?

✓Beta decay involves a neutron converting into a proton and an electron. Since the total number of nucleons (protons + neutrons) is unaffected by simply converting one nucleon type into another, the nucleon number A stays the same, while the proton number Z increases by 1 (as a neutron becomes a proton).

Mastery Test — Answer Key

  1. 14 (27 − 13)
  2. 18 (8 + 10)
  3. 62.5 counts/min — 30÷10=3 half-lives: 500→250→125→62.5
  4. 30 minutes — 1000→500→250→125 is 3 halvings; 90÷3
  5. 90 counts/min (110 − 20)
  6. Alpha particles are relatively large, slow-moving, and carry a large charge, so they interact strongly and very frequently with air molecules, losing their energy (and being absorbed) within a very short distance.
  7. ²³⁸₉₂U → ²³⁴₉₀Th + ⁴₂He (check: 238=234+4 ✓; 92=90+2 ✓)
  8. “Spontaneous” means the decay is not triggered by any external cause and cannot be predicted or influenced from outside the nucleus. “Random” means it is impossible to predict exactly when any individual nucleus will decay, only the overall pattern for a large sample (via half-life).
  9. 480 counts/min — work backward: 30 × 2⁴ = 30 × 16
  10. Any one of: reduce exposure time, increase distance from the source, or use appropriate shielding (e.g. lead) to absorb the radiation.

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