The activity of a radioactive sample is 9750 | vedclass

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Question

(C) The activity of a radioactive sample follows the law $A(t) = A_0 e^{-\lambda t}$.Given: $A_0 = 9750 	ext{ counts/min}$,$A(t) = 975 	ext{ counts/

Vedclass · Accepted Answer

$0.461$

Vedclass · Answer

(C) The activity of a radioactive sample follows the law $A(t) = A_0 e^{-\lambda t}$.Given: $A_0 = 9750 	ext{ counts/min}$,$A(t) = 975 	ext{ counts/min}$,and $t = 5 	ext{ min}$.Substituting the values: $975 = 9750 e^{-5\lambda}$.$\frac{975}{9750} = e^{-5\lambda} \Rightarrow 0.1 = e^{-5\lambda}$.Taking the natural logarithm on both sides: $\ln(0.1) = -5\lambda$.Since $\ln(0.1) = -\ln(10) \approx -2.3026$,we have $-2.3026 = -5\lambda$.$\lambda = \frac{2.3026}{5} = 0.46052 \approx 0.461 	ext{ min}^{-1}$.

The activity of a radioactive sample is $9750$ counts/minute at $t = 0$ and $975$ counts/minute at $t = 5$ minutes. The decay constant is .......... $min^{-1}$.

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