Answer to Question #120030 in Atomic and Nuclear Physics for Priscilla

Question

E. The radioactive nuclide Plutonium-199 has a half-life of 43.0 min. A sample is
prepared that has an initial activity of 7.56 ×1011 Bq.
i. How many Plutonium-199 nuclei are initially present in the sample?
ii. How many are present after 30.8 min?
iii. What is the activity at this time?

Answer 1

Given data

Half life of Plutonium-199  is $T_{1/2}=43.0 min =2580 s$

i)

The Initial number of nuclei is

$N_0=\frac{R_0}{T_{1/2}}=\frac{7.56 ×10^{11} Bq}{2580 s}$

$=2.93*10^8$

ii)

The number of nuclei present after 30.8 min is

$N=N_0e^{-\frac{ln 2t}{T_{1/2}}}$

$=(2.93*10^8)e^{-\frac{ln 2(30.8 min)}{43.0 min}}=1.78*10^8$

iii)

The activity of sample present after 30.8 min is

$R=R_0e^{-\frac{ln 2t}{T_{1/2}}}$

$=(7.56 ×10^{11} Bq)e^{-\frac{ln 2(30.8 min)}{43.0 min}}=4.60*10^{11} Bq$