Question #120030

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?

Expert's answer

Given data

Half life of Plutonium-199  is T1/2=43.0min=2580sT_{1/2}=43.0 min =2580 s

i)

The Initial number of nuclei is

N0=R0T1/2=7.56×1011Bq2580sN_0=\frac{R_0}{T_{1/2}}=\frac{7.56 ×10^{11} Bq}{2580 s}

=2.93108=2.93*10^8

ii)

The number of nuclei present after 30.8 min is

N=N0eln2tT1/2N=N_0e^{-\frac{ln 2t}{T_{1/2}}}

=(2.93108)eln2(30.8min)43.0min=1.78108=(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=R0eln2tT1/2R=R_0e^{-\frac{ln 2t}{T_{1/2}}}

=(7.56×1011Bq)eln2(30.8min)43.0min=4.601011Bq=(7.56 ×10^{11} Bq)e^{-\frac{ln 2(30.8 min)}{43.0 min}}=4.60*10^{11} Bq


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

LATEST TUTORIALS
APPROVED BY CLIENTS