"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 : $\left(\right.$Take $\left.\log _{10} 1.88=0.274\right)$"

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Accepted Answer

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${A_0} = 4250$

$A = 2250 = {A_0}{e^{ - \lambda t}}$

$\Rightarrow {{2250} \over {4250}} = {e^{ - \lambda t}}$

$\Rightarrow \lambda (10) = \ln \left( {{{4250} \over {2250}}} \right)$

$\lambda (10) = 0.636$

$\lambda = 0.063$

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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 :

$\left(\right.$Take $\left.\log _{10} 1.88=0.274\right)$

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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 : $\left(\right.$Take $\left.\log _{10} 1.88=0.274\right)$

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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 :

$\left(\right.$Take $\left.\log _{10} 1.88=0.274\right)$