A radioactive isotope has an initial activity of 800 counts per second. If its half-life is 4.0 hours, what will its activity be after 12.0 hours?
Answer and explanation
Correct answer: A. 100 counts per second
- Step 1: Determine number of half-lives elapsed: n = t / T_half = 12.0 / 4.0 = 3 half-lives. Step 2. Apply decay fraction: A = A0 × (1/2)ⁿ = 800 × (1/2)³ = 800 / 8 = 100 counts per second. Answer: 100 counts per second. Why not the others: 200 counts/s is after 2 half-lives (8 h); 50 counts/s is after 4 half-lives (16 h); 400 counts/s is after 1 half-life (4 h).
Practise more Electricity, Magnetism and Modern Physics.
Free questions with an explanation for every one, and full timed Unified Tertiary Matriculation Examination mock tests.
More Electricity, Magnetism and Modern Physics questions
- Calculate the electrostatic force of attraction between two point charges of +2.0 μC and -4.0 μC separated by a distance of 0.20 m in a vacu…
- Determine the electric field intensity at a distance of 0.30 m from an isolated positive point charge of 5.0 nC in air. (Take k = 9.0 × 10⁹…
- A parallel-plate capacitor has plates of area 0.040 m² separated by a vacuum gap of 2.0 mm. Calculate its capacitance. (Permittivity of free…
- Two capacitors of capacitances 4.0 μF and 6.0 μF are connected in series across a direct current voltage supply. What is the equivalent capa…
- A 10 μF capacitor is charged to a potential difference of 200 V. Calculate the total electrostatic energy stored in the capacitor.
- A chemical cell of electromotive force (emf) 1.50 V and internal resistance 0.50 Ω is connected across an external resistor of 2.50 Ω. Calcu…