4. A radioactive isotope has a half-life of 76.0 min. A sample is prepared that has an initial activity of 16.0E10 Bq.
a. How many radioactive nuclei are initially present in the sample? Answer:1.05E nuclei
b. How many are present after 76.0 min? Answer: 5.25E14 nuclei
d. How many radioactive nuclei are present 152.0 min after the sample was first prepared? Answer: 2.62E14 nuclei
e. What is the activity after 152.0 min? Express answer in Bq.Answer: 4.0E10 Bq
Law of radioactive decay:
N = No e-kt
Half-life, t1/2 = 0.693 / k
Thus, 0.693 / k = 76 * 60 or k = 1.5197E-4
Law of radioactive decay:
dN/dt = -kN = -k(m/M)N
or 16.0E10 Bq = - 1.5197E-4 * N
or N = 10.5E14 ignoring negative sign
After 76 min,
N = 10.5E14 * e-1.5197E-4*76*60
= 10.5E14 * e-0.69298
= 10.5E14 * 0.5
= 5.25E14
After 152 min,
N = 10.5E14 * e-1.5197E-4*152*60
= 10.5E14 * e-1.386
= 10.5E14 * 0.25
= 2.625E14
Activity after 152 min = kN = 2.625E14 * 1.5197E-4
= 3.989E10