Physics – 5.2.4 Half-life | e-Consult

Solution:

1. The relationship between activity and time is given by: A(t) = A₀e^-kt, where A₀ is the initial activity, k is the decay constant, and t is time.
2. We are given that A(200) = A₀/3 and t = 200 s. Substituting these values: A₀/3 = A₀e^-200k
3. Divide both sides by A₀: 1/3 = e^-200k
4. Take the natural logarithm of both sides: ln(1/3) = -200k => -1.099 = -200k
5. Solve for k: k = -1.099 / -200 = 0.005495 s^-1
6. The half-life (T_1/2) is given by: T_1/2 = ln(2) / k
7. Substitute the value of k: T_1/2 = 0.693 / 0.005495 = 126 s
8. Therefore, the half-life of the radioactive substance is 126 seconds.