The disintegration rate of a certain radioact | vedclass

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Question

(C) The disintegration rate $A$ follows the law $A = A_0 e^{-\lambda t}$.At $t = 0$,$A_0 = 4250 \, 	ext{dpm}$.At $t = 10 \, 	ext{min}$,$A = 2250 \,

Vedclass · Accepted Answer

$0.063$

Vedclass · Answer

(C) The disintegration rate $A$ follows the law $A = A_0 e^{-\lambda t}$.At $t = 0$,$A_0 = 4250 \, 	ext{dpm}$.At $t = 10 \, 	ext{min}$,$A = 2250 \, 	ext{dpm}$.Substituting these values into the equation:$2250 = 4250 e^{-\lambda (10)}$$e^{-10\lambda} = \frac{2250}{4250} = \frac{45}{85} = \frac{9}{17}$Taking the natural logarithm on both sides:$-10\lambda = \ln\left(\frac{9}{17}ight)$$10\lambda = \ln\left(\frac{17}{9}ight) \approx \ln(1.888)$$10\lambda \approx 0.6356$$\lambda \approx 0.06356 \, \min^{-1}$.Thus,the approximate decay constant is $0.063 \, \min^{-1}$.

The disintegration rate of a certain radioactive sample at any instant is $4250$ disintegrations per minute. $10$ minutes later,the rate becomes $2250$ disintegrations per minute. The approximate decay constant is $......... \min^{-1}$.

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