Nuclear Physics And Radioactivity Question 437
Question: For a radioactive sample the counting rate changes from 6520 counts/minute to 3260 counts minute in 2 minutes. Determine the decay constant.
Options:
A) 1.78 per sec
B) 0.78 per sec
C) 2.78 per sec
D) 5.78 per sec
Show Answer
Answer:
Correct Answer: D
Solution:
[d] at time $ t=0; $ $ A_{0}=\frac{dN_{0}}{dt} $ $ \frac{A}{A_{0}}=\frac{dN/dt}{dN_{0}/dt}=\frac{3260}{6520} $ or $ \lambda =\frac{2.303}{t}\log \frac{A_{0}}{A}=\frac{2.303}{2\times 60}\log 2 $ $ =5.78per\sec $
BETA
