The Physics Question Bank

F: T_½ = ln 2 ÷ λ
I: 8.0 = ln 2 ÷ λ
F: λ = ln 2 ÷ 8.0 = 0.6931… ÷ 8.0
A: 0.08664… day⁻¹ = 0.087 day⁻¹ (2 s.f.)

F: T_½ = ln 2 ÷ λ
I: T_½ = 0.6931… ÷ 3.2 × 10⁻⁵
F: T_½ is already the subject
A: 21 660… s = 2.2 × 10⁴ s (2 s.f.); 21 660… ÷ 3600 = 6.01… h = 6.0 h (2 s.f.)

Step 1 – number of half-lives
n = t ÷ T_½ = 24 ÷ 6.0 = 4
F: A = A₀ ÷ 2ⁿ
I: A = 6.4 × 10⁵ ÷ 2⁴ = 6.4 × 10⁵ ÷ 16
F: A is already the subject
A: 4.0 × 10⁴ Bq; fraction remaining = 1 ÷ 2⁴ = 1/16 (0.0625)
Watch out: 4 half-lives does not mean ¼ is left. Each half-life halves what is left, so it is ½ × ½ × ½ × ½ = 1/16.

Convert first: T_½ = 432 × 3.15 × 10⁷ = 1.3608 × 10¹⁰ s
F: T_½ = ln 2 ÷ λ
I: 1.3608 × 10¹⁰ = 0.6931… ÷ λ
F: λ = 0.6931… ÷ 1.3608 × 10¹⁰
A: 5.093… × 10⁻¹¹ s⁻¹ = 5.09 × 10⁻¹¹ s⁻¹ (3 s.f.)
Watch out: working out ln 2 ÷ 432 gives λ in year⁻¹, not s⁻¹. Convert the half-life to seconds first if you need λ in s⁻¹.

Step 1 – number of half-lives
12.5% = 0.125 = 1/8 = 1 ÷ 2³, so n = 3
Step 2 – half-life
T_½ = t ÷ n = 15 ÷ 3 = 5.0 days
Step 3 – decay constant
F: T_½ = ln 2 ÷ λ
I: 5.0 = 0.6931… ÷ λ
F: λ = 0.6931… ÷ 5.0
A: 0.1386… day⁻¹ = 0.14 day⁻¹ (2 s.f.)

Step 1 – (a) fraction remaining
n = 24 ÷ 6.0 = 4, so fraction remaining = 1 ÷ 2⁴ = 1/16 = 0.0625
A: 6.25% (6.3% to 2 s.f.)
Step 2 – (b) choice of half-life
6.0 h is long enough for the tracer to be prepared, injected and imaged while its activity is still high enough to detect
it is short enough that the activity in the patient falls quickly, so the total dose received is small; after a day only about 6% remains
with T_½ = 30 s most of the activity would be gone before imaging; with T_½ = 30 years the patient would stay radioactive for the rest of their life
gamma rays are also weakly ionising and penetrate the body, so they reach the detector outside
Watch out: the answer must link the half-life to both detection and patient dose. “It is safe” on its own gains no marks.

Step 1 – (a) decay constant
λ = ln 2 ÷ T_½ = 0.6931… ÷ 432 = 1.604… × 10⁻³ year⁻¹
Step 2 – (a) fraction remaining
F: A = A₀e^−λt
I: A ÷ A₀ = e^{−1.604… × 10−3 × 10} = e^{−0.01604…}
A: 0.9840… = 98.4% (3 s.f.)
Step 3 – (b) explanation
the long half-life means the activity (and so the ionisation current in the alarm) stays almost constant over the life of the alarm, so it does not need replacing
alpha particles are strongly ionising, so they ionise the air between the plates, but they have a range of only a few cm in air and are stopped by smoke particles and the plastic case, so the alarm is safe
Watch out: 10 years is not a whole number of half-lives, so 1 ÷ 2ⁿ needs n = 10 ÷ 432. Using A = A₀e^−λt is clearer.

Step 1 – correct for background
corrected count rate at t = 0: 520 − 40 = 480 counts per minute
corrected count rate at t = 10 min: 160 − 40 = 120 counts per minute
Step 2 – number of half-lives
480 ÷ 120 = 4 = 2², so n = 2 half-lives in 10 min
F: n = t ÷ T_½
I: 2 = 10 ÷ T_½
F: T_½ = 10 ÷ 2
A: 5.0 min
Watch out: without subtracting background the ratio is 520 ÷ 160 = 3.25, which wrongly gives a half-life of about 5.9 min.

Convert first: λ = ln 2 ÷ 6.0 = 0.1155… h⁻¹ (MBq can stay as MBq because only the ratio is needed)
F: A = A₀e^−λt
I: 5.0 = 400 × e^{−0.1155… × t}
F: t = ln(400 ÷ 5.0) ÷ 0.1155… = ln 80 ÷ 0.1155… = 4.382… ÷ 0.1155…
A: t = 37.93… h = 38 h (2 s.f.)
Check: after 6 half-lives (36 h) A = 400 ÷ 2⁶ = 6.25 MBq, still above 5.0 MBq; after 7 half-lives (42 h) A = 400 ÷ 2⁷ = 3.125 MBq, below 5.0 MBq, so the answer lies between 36 h and 42 h

Step 1 – correct for background
corrected count rate at t = 0: 524 − 24 = 500 counts per minute
corrected count rate at the later time: 60 − 24 = 36 counts per minute
Step 2 – decay constant
λ = ln 2 ÷ T_½ = 0.6931… ÷ 2.6 = 0.2665… min⁻¹
Step 3 – time
F: C = C₀e^−λt (corrected count rate is proportional to activity)
I: 36 = 500 × e^{−0.2665… × t}
F: t = ln(500 ÷ 36) ÷ 0.2665… = 2.631… ÷ 0.2665…
A: t = 9.869… min = 9.9 min (2 s.f.)
Watch out: the background does not decay. Using 524 and 60 without correcting gives about 8.1 min, which is wrong.