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(C) The radioactive decay law states that the amount remaining after $n$ half-lives is given by $N = N_0 (1/2)^n$.To reduce the substance to one-fourt

Vedclass · Accepted Answer

$80 \, years, 0.0173 \, year^{-1}$

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(C) The radioactive decay law states that the amount remaining after $n$ half-lives is given by $N = N_0 (1/2)^n$.To reduce the substance to one-fourth of its original amount $(N = N_0/4)$, we have $(1/2)^n = 1/4$, which implies $n = 2$.Since one half-life $(T_{1/2})$ is $40 \, years$, the total time taken is $t = n 	imes T_{1/2} = 2 	imes 40 = 80 \, years$.The decay constant $\lambda$ is given by the formula $\lambda = \frac{0.693}{T_{1/2}}$.Substituting the given value: $\lambda = \frac{0.693}{40} = 0.0173 \, year^{-1}$.

The half-life of a radioactive substance is $40 \, years$. How long will it take to reduce to one-fourth of its original amount, and what is the value of the decay constant?

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