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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

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N = No e-kt

Half-life, t1/2 = 0.693 / k

Thus, 0.693 / k = 76 * 60 or k = 1.5197E-4

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

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