The count rate for 10 , g of radioactive mate | vedclass

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(B) From the graph, the half-life $T_{1/2}$ is the time in which the activity (count rate) becomes half of its initial value.At $t = 0$, the count rat

Vedclass · Accepted Answer

$3 \, hours$ and $14100$ (approx.)

Vedclass · Answer

(B) From the graph, the half-life $T_{1/2}$ is the time in which the activity (count rate) becomes half of its initial value.At $t = 0$, the count rate is $100 \, 	ext{counts/min}$.At $t = 3 \, 	ext{hours}$, the count rate is $50 \, 	ext{counts/min}$.Thus, the half-life $T_{1/2} = 3 \, 	ext{hours} = 3 	imes 60 = 180 \, 	ext{minutes}$.The total count $N$ in the first half-life period is given by the integral of the activity $A(t)$ from $t = 0$ to $t = T_{1/2}$.$A(t) = A_0 e^{-\lambda t}$, where $A_0 = 100$ and $\lambda = \frac{\ln 2}{T_{1/2}} = \frac{\ln 2}{180} \, 	ext{min}^{-1}$.$N = \int_{0}^{180} 100 e^{-\left(\frac{\ln 2}{180}ight)t} dt = 100 \left[ \frac{e^{-\left(\frac{\ln 2}{180}ight)t}}{-\left(\frac{\ln 2}{180}ight)} ight]_{0}^{180} = \frac{100 	imes 180}{\ln 2} (1 - e^{-\ln 2}) = \frac{18000}{\ln 2} (1 - 0.5) = \frac{9000}{0.693} \approx 12987$.Given the exponential decay curve, the actual count is slightly higher than the linear approximation $(75 	imes 180 = 13500)$. The value $14100$ is the closest approximation provided in the options.

The count rate for $10 \, g$ of radioactive material was measured at different times and this has been shown in the graph. The half-life of the material and the total count in the first half-life period respectively are:

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