a. The rate of decay = 26,880 dpm (decays/min)
b.
k = 0.693/t1/2
Solve for k and substitute into the below equation.
ln(No/N) = kt
No = 26880
N = ?
k from above
t = 265 days
c.
1 Bq = 1 dps
Your sample is 26,880 dpm so
26,880 decays/min x (1 min/60 sec) = about 448 dps.
ln(No/N) = kt
No = 448
N = 112
k from above
Solve for t in days.
d.
ln(No/N) = kt
No = ?
N = 26,880
k from above
t = 106
e. see d.
Beryllium-7 has a half-life of 53 D. A sample was observed for one minute and there were 26,880 decays.
a.) what is the activity level of the sample?
b.) what will the activity level of the sample be after 265 d?
c.) after how many days will be activity level of the sample be 112 Bq?
d.) what was activity level 106 d before the sample was observed?
e.) how many days earlier was activity level 8 times greater than the observed level?
1 answer
1 answer