Radioactive decay is characterized by the equation In(N_t/N_0) = -kt where N_0 i

Question

Radioactive decay is characterized by the equation In(N_t/N_0) = -kt where N_0 is the initial amount, N_t is the amount remaining at time t, and k is the rate constant. The half-life (t_1/2) is the time required for the number of radioactive nuclei in a sample to drop to half of its initial value. It is defined as t_1/2 = In 2/k where k is the decay constant. Phosphorus-32 () decays by beta emission to form sulfur-32 (). How many half-lives have passed in the reaction shown here? The half-life of phosphorus-32 is 14.26 days. Calculate its decay constant.

Explanation / Answer

A)

we know that

ln ( N/No) = -kt

from the figure we can see that

inital number of atoms of phosphorus (No) = 24

final number of atoms of phosphorus (N) = 3

so

ln ( 3/24) = -kt

kt = 2.07944

now

k = ln2 / t1/2

so

t x ln 2 / t1/2 = 2.07944

t = 3 x t1/2

so

three half lives have passed

so

the answer is 3

B)

we know that

decay constant (k) = ln2 / t1/2

given

t1/2 = 14.26

so

k = ln2 / 14.26

k = 0.0486 -1

so

the decay constant is 0.0486 days-1

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