# 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) Option 2) Option 3) Option 4)

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

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

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

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

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

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)

This is an incorrect option.

Option 2)

This is the correct option.

Option 3)

This is an incorrect option.

Option 4)

This is an incorrect option.

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