Half-life of a radioactive substance is 20 mi | vedclass

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Question

(b) $\lambda = \frac{{0.693}}{{{T_{1/2}}}} = \frac{{0.693}}{{20}} = 0.03465$

Now time of decay $t = \frac{{2.303}}{\lambda }\log \frac{{{N_0}}}{N}$

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$20 $

Vedclass · Answer

(b) $\lambda = \frac{{0.693}}{{{T_{1/2}}}} = \frac{{0.693}}{{20}} = 0.03465$

Now time of decay $t = \frac{{2.303}}{\lambda }\log \frac{{{N_0}}}{N}$

$ \Rightarrow {t_1} = \frac{{2.303}}{{0.03465}}\log \frac{{100}}{{67}} = 11.6\, min $

and ${t_2} = \frac{{2.303}}{{0.03465}}\log \frac{{100}}{{33}} = 32\,min$

Thus time difference between points of time

$= t_1 -t_2 =32 -11.6 = 20.4\, min \, \approx \, 20\, min.$

Half-life of a radioactive substance is $20$ minutes. Difference between points of time when it is $33\%$ disintegrated and $67\%$ disintegrated is approximately ........... $min$

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