A sample of radioactive material A, that has an activity of 10 mCi (1 Ci = 3.7 x 1010 decays/s), has twice the number of nuclei as another sample of a different radioactive material B which has an activity of 20 mCi. The correct choices for half-lives of A and B would then by respectively :

  • Option 1)

    5 days and 10 days

  • Option 2)

    10 days and 40 days

  • Option 3)

    20 days and 5 days

  • Option 4)

    20 days and 10 days

Answers (1)
A admin

 

Law of radioactivity -

-\frac{dN}{dt}= \lambda N

- wherein

Ratio of disintegration is propotional to number of nuclei         

\lambda= disintegration constant

 

Activity A = \lambda N

For A , A = 2 No \lambda_{A} = 10

For B , A =  No \lambda_{B} = 20

 

\frac{10}{20} = \frac{2 \lambda_{A}}{\lambda_{B}}

\lambda_{B} = 4 \lambda_{A}

t_{\frac{1}{2} }= \frac{ln 2 }{\lambda}

\frac{t_{\frac{1}{2}A }}{t_{\frac{1}{2}B }}= \frac{\lambda_{B}}{\lambda_{A}} = \frac{\lambda_{B}}{\frac{\lambda_{B}}{4}} = 4

t_{\frac{1}{2}A } = 4 t_{\frac{1}{2}B }

So option 20 days and 5 days satisfies this condition .

 


Option 1)

5 days and 10 days

Option 2)

10 days and 40 days

Option 3)

20 days and 5 days

Option 4)

20 days and 10 days

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