Nuclear Physics And Radioactivity Question 480
Question: The count rate from a radioactive sample falls from $ 4.0\times 10^{6} $ per second to $ 1\times 10^{6} $ per second in 20 hour. What will be the count rate, 100 hour after the beginning?
Options:
A) $ 3.91\times 10^{3}se{{c}^{-1}} $
B) $ 5.81\times 10^{4}se{{c}^{-1}} $
C) $ 6.22\times 10^{5}se{{c}^{-1}} $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
[a] If $ A_{0} $ is the initial activity of radioactive sample then activity at any time $ A=A_{0}{{e}^{-\lambda t}} $ or $ 1\times 10^{6}=4\times 10^{6}{{e}^{-\lambda \times 20}} $ or $ {{e}^{-20\lambda }}=\frac{1}{4} $ The count rate after 100 hour is given by $ A’=A_{0}{{e}^{-\lambda \times 100}}=A_{0}{{e}^{-100\lambda }}=A_{0}{{[{{e}^{-20\lambda }}]}^{5}} $ $ =4\times 10^{6}{{[ \frac{1}{4} ]}^{5}}=3.91\times 10^{3} $ per second
BETA
