Atoms And Nuclei Question 367

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:

  • 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 $



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