The decay rate of radioactive material at any | vedclass

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Question

(C) Let $x$ be the mass of the material at time $t$. The rate of decay is given by $\frac{dx}{dt} = -kx$. Integrating this,we get $\ln(x) = -kt + C$.

Vedclass · Accepted Answer

$\frac{16}{3} 	ext{ grams}$

Vedclass · Answer

(C) Let $x$ be the mass of the material at time $t$. The rate of decay is given by $\frac{dx}{dt} = -kx$. Integrating this,we get $\ln(x) = -kt + C$. At $t=0, x=27$,so $C = \ln(27)$. Thus,$\ln(x) = -kt + \ln(27)$. At $t=3, x=8$,so $\ln(8) = -3k + \ln(27)$,which gives $3k = \ln(27) - \ln(8) = \ln(\frac{27}{8})$. Therefore,$k = \frac{1}{3} \ln(\frac{27}{8}) = \ln((\frac{27}{8})^{1/3}) = \ln(\frac{3}{2})$. We need to find the mass after one more hour,i.e.,at $t=4$. $\ln(x) = -4 \ln(\frac{3}{2}) + \ln(27) = \ln((\frac{3}{2})^{-4}) + \ln(27) = \ln(\frac{16}{81} 	imes 27) = \ln(\frac{16}{3})$. Therefore,$x = \frac{16}{3} 	ext{ grams}$.

The decay rate of radioactive material at any time $t$ is proportional to its mass at that time. The mass is $27 \text{ grams}$ when $t=0$. After $3 \text{ hours}$,it was found that $8 \text{ grams}$ are left. Then the substance left after one more hour is

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