10 , g of radioactive material of half-life 1 | vedclass

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(C) The amount of radioactive material remaining after time $t$ is given by the formula $N = N_0 \left( \frac{1}{2} \right)^{t/T}$, where $N_0$ is the

Vedclass · Accepted Answer

$6.04$

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(C) The amount of radioactive material remaining after time $t$ is given by the formula $N = N_0 \left( \frac{1}{2} ight)^{t/T}$, where $N_0$ is the initial amount, $t$ is the time elapsed, and $T$ is the half-life.Given: $N_0 = 10 \, g$, $t = 20 \, 	ext{years}$, $T = 15 \, 	ext{years}$.Substituting the values: $N = 10 	imes (2)^{-20/15} = 10 	imes (2)^{-4/3} = 10 	imes (2)^{-1.333}$.Calculating the value: $2^{1.333} \approx 2.5198$.So, $N = 10 / 2.5198 \approx 3.968 \, g$.The amount of disintegrated (decayed) material is $N_{	ext{decayed}} = N_0 - N$.$N_{	ext{decayed}} = 10 - 3.968 = 6.032 \, g$.Rounding to the nearest provided option, the disintegrated material is $6.04 \, g$.

$10 \, g$ of radioactive material of half-life $15 \, \text{years}$ is kept in store for $20 \, \text{years}$. The disintegrated material is ............ $g$.

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