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The decay constant of a radioisotope is \lambda . if A_{1} and A_{2} are its activities at times t_{1} and t_{2} respectively, the number of nuclei which have decayed during the time\left ( t_{1} \right-_{t_{2}} ) :

  • Option 1)

    \lambda \left ( A_{1} \right-A_{2} )

  • Option 2)

    A_{1}t_{1}-A_{2}t_{2}

  • Option 3)

    A_{1}-A_{2}

  • Option 4)

    \left ( A_{1} \right-A_{2} )/\lambda

 

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

 

 

 

Law of radioactivity -

-frac{dN}{dt}= lambda N

- wherein

Ratio of disintegration is propotional to number of nuclei         

lambda= disintegration constant

 

 

A_{1}=A_{o}\cdot e^{-\lambda t_{1}} \: \: \Rightarrow N_{1}=\frac{A_{1}}{\lambda }=\frac{A_{o}}{\lambda }\cdot e^{-\lambda t_{2}}

A_{2}=A_{o}\cdot e^{-\lambda t_{2}} \: \: \Rightarrow N_{2}=\frac{A_{2}}{\lambda }=\frac{A_{o}}{\lambda }\cdot e^{-\lambda t_{1}}

Number of nuclei decayed between t1 and t2 = N1-N2= \frac{A_{1}-A_{2}}{\lambda }


Option 1)

\lambda \left ( A_{1} \right-A_{2} )

Incorrect

Option 2)

A_{1}t_{1}-A_{2}t_{2}

Incorrect

Option 3)

A_{1}-A_{2}

Incorrect

Option 4)

\left ( A_{1} \right-A_{2} )/\lambda

Correct

Posted by

divya.saini

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