Question 212: Atoms & Nuclei

Question

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

Accepted Answer

$${A_0} = 4250$$
$$A = 2250 = {A_0}{e^{ - \lambda t}}$$
$$ \Rightarrow {{2250} \over {4250}} = {e^{ - \lambda t}}$$
$$ \Rightarrow \lambda (10) = \ln \left( {{{4250} \over {2250}}} \right)$$
$$\lambda (10) = 0.636$$
$$\lambda  = 0.063$$

Answer

$$0.02 \min ^{-1}$$

Answer

$$2.7 \min ^{-1}$$

Answer

$$0.063 \min ^{-1}$$

Answer

$$6.3 \min ^{-1}$$

Atoms & Nuclei

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)

Choose the correct answer: