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The activity of a radioactive sample is measured as N_{0} counts per minute at t=0 and   N_{0}/e counts per minute at t=5 minutes. The time (in minutes) at which the actvity reduces to half its value is

  • Option 1)

    log_{e} 2/5

  • Option 2)

    \frac{5}{log_{e}2}

  • Option 3)

    5 \log_{10}2

  • Option 4)

    5log_{e}2

 

Answers (1)

best_answer

As we learnt in 

Number of nuclei after disintegration -

N=N_{0}e^{-lambda t} or A=A_{0}e^{-lambda t}

- wherein

Number of nucleor activity at a time is exponentional function

 

 

N=N_{o}\cdot e^{-\lambda t}

at t=5 min, N=\frac{N_{o}}{e} \: = N_{o}\cdot e^{-\lambda t}

\Rightarrow t=\frac{1}{\lambda} \: \: \: \: or \: \: \lambda=\frac{1}{5min}

t_{\frac{1}{2}}=\frac{ln^{2}}{\lambda}= \: \: 5.\log_{e}2 \: min


Option 1)

log_{e} 2/5

Incorrect

Option 2)

\frac{5}{log_{e}2}

Incorrect

Option 3)

5 \log_{10}2

Incorrect

Option 4)

5log_{e}2

Correct

Posted by

prateek

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