A radioactive nucleus A with a half life T, decays into a nucleus B.  At t=0, there is no nucleus B.  At sometime t, the ratio of the number of B to that of A is 0.3.  Then, t is given by :

  • Option 1)

    t= \frac{T}{2}\: \: \frac{\log 2}{\log 1.3}

  • Option 2)

    t= T\frac{\log 1.3}{\log 2}

  • Option 3)

    t= T\log \left ( 1.3 \right )

  • Option 4)

    t= \frac{T}{\log 1.3}

 

Answers (1)

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

 

 Number of nucleus of A at any time t is 

N_{A}=N_{0}.E^{-\lambda t}

NB= Number of nucleus of A that decayed tinto B

=N_{0}\left(1-e^{-\lambda t} \right )

\frac{N_{B}}{N_}=\frac{1-e^{-\lambda t}}{e^{-\lambda t}}=0.3

\therefore\ \; 1-e^{-\lambda t}=0.3e^{-\lambda t}\ or\; 1.3e^{-\lambda t}=1

1-e^{-\lambda t}=\frac{1}{1.3}

Taking log

\lambda t=log_{e}1.3

t.\left(\frac{log_{e}\ {2}}{T} \right )=log_{e}\ 1.3

t=T\frac{log_{e}1.3}{log_{e}2}

Correct answer is 2


Option 1)

t= \frac{T}{2}\: \: \frac{\log 2}{\log 1.3}

This is an incorrect option.

Option 2)

t= T\frac{\log 1.3}{\log 2}

This is the correct option.

Option 3)

t= T\log \left ( 1.3 \right )

This is an incorrect option.

Option 4)

t= \frac{T}{\log 1.3}

This is an incorrect option.

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