The activity of a radioisotope is 3000 counts per minute at one time and 2736 counts per minute 48 hours later. What is the half-life of the radioisotope??

Question

a) 831hr
b)521hr
c)361 hr
d)1.44hr

Answer 1

ln(No/N) = kt
No = 3000
N = 2736
k = ?
t = 48 hrs.
Solve for k.

Then k = 0.693/t_1/2
Substitute k from above and solve to t1/2

Answer 2

Dr. Bob,

Where are you getting the 0.693 from?

Answer 3

log base 3 of 1 is .693

Answer 4

typo
log base e (as in ln) of 2 is .693

Answer 5

Bob Pursley is right. More explicitly, it comes from this.
If we start with 100 counts and drop to 50 counts (which is 1 half life or t1/2), then
ln(No/N) = kt
ln(100/50) = kt_1/2
ln 2 = kt_1/2
0.693 = kt_1/2
and k = 0.693/t_1/2

Answer 6

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