The decay constant of a radioactive substance | vedclass

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Question

(D) Given decay constant $\lambda = 0.173 \, (years)^{-1}.$$1$. Half-life $T_{1/2} = \frac{\ln 2}{\lambda} = \frac{0.693}{0.173} \approx 4 \, years.$$

Vedclass · Accepted Answer

$A$ and $C$ both.

Vedclass · Answer

(D) Given decay constant $\lambda = 0.173 \, (years)^{-1}.$$1$. Half-life $T_{1/2} = \frac{\ln 2}{\lambda} = \frac{0.693}{0.173} \approx 4 \, years.$$2$. For time $t = 1/\lambda = 1/0.173 \, years,$ the remaining amount is $N = N_0 e^{-\lambda t} = N_0 e^{-1} \approx 0.37 N_0.$Thus,the decayed amount is $N_0 - 0.37 N_0 = 0.63 N_0,$ which is $63\%.$ So,option $A$ is correct.$3$. After $8 \, years,$ which is $2 	imes T_{1/2},$ the remaining amount is $N = N_0 (1/2)^2 = N_0/4.$ So,option $C$ is correct.Therefore,both $A$ and $C$ are correct.

The decay constant of a radioactive substance is $0.173 \, (years)^{-1}.$ Therefore:

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