Question
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??
a) 831hr
b)521hr
c)361 hr
d)1.44hr
a) 831hr
b)521hr
c)361 hr
d)1.44hr
Answers
DrBob222
ln(No/N) = kt
No = 3000
N = 2736
k = ?
t = 48 hrs.
Solve for k.
Then k = 0.693/t<sub>1/2</sub>
Substitute k from above and solve to t</sub>1/2</sub>
No = 3000
N = 2736
k = ?
t = 48 hrs.
Solve for k.
Then k = 0.693/t<sub>1/2</sub>
Substitute k from above and solve to t</sub>1/2</sub>
Michelle
Dr. Bob,
Where are you getting the 0.693 from?
Where are you getting the 0.693 from?
bobpursley
log base 3 of 1 is .693
bobpursley
typo
log base e (as in ln) of 2 is .693
log base e (as in ln) of 2 is .693
DrBob222
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 t</sub>1/2</sub>), then
ln(No/N) = kt
ln(100/50) = kt<sub>1/2</sub>
ln 2 = kt<sub>1/2</sub>
0.693 = kt<sub>1/2</sub>
and k = 0.693/t<sub>1/2</sub>
If we start with 100 counts and drop to 50 counts (which is 1 half life or t</sub>1/2</sub>), then
ln(No/N) = kt
ln(100/50) = kt<sub>1/2</sub>
ln 2 = kt<sub>1/2</sub>
0.693 = kt<sub>1/2</sub>
and k = 0.693/t<sub>1/2</sub>
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