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  • Answer please! A certain radioactive material is known to decay at a rate propotional to the amount present. If initially there is 50 mg of material present and after two hours, it is observed that material has lost 10% of its mass. Then the total ti

A certain radioactive material is known to decay at a rate propotional to the amount present. If initially there is 50 mg of material present and after two hours, it is observed that material has lost 10% of its mass. Then the total time after which material is lost 50% of its original amount will be (in hours)

  • Option 1)

    \log_{\;0.9}(0.25)

  • Option 2)

    \log_{\;10}(0.25)

  • Option 3)

    \log_{\;0.25}(0.9)

  • Option 4)

    10

 

Answers (1)

best_answer

As we have learnt,

 

Growth and Decay Problems -

We assume that rate of change of amount of substance is proportional to the amount of substance present

-

 

 

Let at time t, and N be the amout of material at that time, So

 \frac{\mathrm d N}{\mathrm dx} = -kN \\*\Rightarrow \frac{\mathrm d N }{N} = -kdt

So, on integrating we get,

\ln N = -kt + c

Intially Material is 50 mg at t = 0

\Rightarrow c = \ln 50

At t =2, N = 45, so

45= 50 e^{-k\cdot 2}\Rightarrow e^{-2k} = \frac{9}{10}\Rightarrow -2k = \ln \frac{9}{10}\Rightarrow k = -\frac{1}{2}\ln \frac{9}{10}

We need to find out t, when N = 25

25= 50\cdot e^{\frac{1}{2}\ln \frac{9}{10}\times t} \Rightarrow \frac{1}{2} = e^{\frac{1}{2}\ln \frac{9}{10}\cdot t} \Rightarrow \frac{1}{2}\ln \frac{9}{10}\cdot t = \ln (0.5) \\*\Rightarrow t = \frac{2\ln(0.5)}{\ln(.9)} = \log_{0.9}(.25)

 


Option 1)

\log_{\;0.9}(0.25)

Option 2)

\log_{\;10}(0.25)

Option 3)

\log_{\;0.25}(0.9)

Option 4)

10

Posted by

Himanshu

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