Physics – 5.2.4 Half-life | e-Consult

Solution:

1. We know that A(t) = A₀e^-kt. We are given A₀ = 1000 and we can use the data from the decay curve to find k. Using t = 100 s and A(100) = 315.5: 315.5 = 1000e^-100k
2. Divide both sides by 1000: 0.3155 = e^-100k
3. Take the natural logarithm of both sides: ln(0.3155) = -100k => -1.152 = -100k
4. Solve for k: k = -1.152 / -100 = 0.01152 s^-1
5. The half-life (T_1/2) is given by: T_1/2 = ln(2) / k
6. Substitute the value of k: T_1/2 = 0.693 / 0.01152 = 60.1 s
7. Therefore, the half-life of the isotope is approximately 60.1 seconds.