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(D) The mass of a radioactive substance at time $t$ is given by $M = M_0 e^{-\lambda t}$.Here,$M_0 = 10 \, gm$ and the time $t$ is equal to two mean l

Vedclass · Accepted Answer

$1.35$

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(D) The mass of a radioactive substance at time $t$ is given by $M = M_0 e^{-\lambda t}$.Here,$M_0 = 10 \, gm$ and the time $t$ is equal to two mean lifetimes,i.e.,$t = 2	au$.Since the mean lifetime $	au = \frac{1}{\lambda}$,we have $t = \frac{2}{\lambda}$.Substituting these values into the decay equation:$M = 10 \cdot e^{-\lambda \left( \frac{2}{\lambda} ight)}$$M = 10 \cdot e^{-2}$$M = 10 \cdot \left( \frac{1}{e^2} ight) \approx 10 \cdot \left( \frac{1}{7.389} ight) \approx 1.353 \, gm$.Rounding to the nearest option,the remaining mass is approximately $1.35 \, gm$.

At time $t = 0$,a radioactive element has a mass of $10 \, gm$. What mass in $gm$ will remain after two mean lifetimes?

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