The count rate of 10,g of radioactive materia | vedclass

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(B) $1$. To find the half-life,observe the graph. The initial count rate at $t = 0$ is $100$. The half-life is the time taken for the count rate to be

Vedclass · Accepted Answer

$3\,h, 14000$

Vedclass · Answer

(B) $1$. To find the half-life,observe the graph. The initial count rate at $t = 0$ is $100$. The half-life is the time taken for the count rate to become half of its initial value,i.e.,$50$. From the graph,at a count rate of $50$,the time is $3\,h$. Thus,the half-life $T_{1/2} = 3\,h$.$2$. To find the total counts in the first half-life period ($0$ to $3\,h$),we calculate the area under the curve between $t = 0$ and $t = 3\,h$.$3$. Each small square on the graph represents a count rate of $10$ units on the y-axis and $1\,h$ on the x-axis. However,the question specifies $10\,g$ of material. The area under the curve represents the total number of counts. By counting the number of small squares under the curve from $t = 0$ to $t = 3\,h$,we find approximately $24$ squares.$4$. Each square represents $10 	imes 60 = 600$ counts (assuming the rate is per minute,so $10 	ext{ counts/min} 	imes 60 	ext{ min} = 600 	ext{ counts/hour}$). $5$. Total counts $\approx 24 	imes 600 = 14400$.

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

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