A radioactive isotope has a half-life of 10 , | vedclass

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(D) Given: Half-life $(t_{1/2})$ = $10 \, 	ext{years}$,Total time $(t)$ = $20 \, 	ext{years}$.
Number of half-lives $(n) = \frac{t}{t_{1/2}} = \fra

Vedclass · Accepted Answer

$25$

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(D) Given: Half-life $(t_{1/2})$ = $10 \, 	ext{years}$,Total time $(t)$ = $20 \, 	ext{years}$.
Number of half-lives $(n) = \frac{t}{t_{1/2}} = \frac{20}{10} = 2$.
The remaining amount $(N)$ is given by the formula $N = N_0 	imes (\frac{1}{2})^n$.
Substituting the values: $N = N_0 	imes (\frac{1}{2})^2 = N_0 	imes \frac{1}{4} = 0.25 \, N_0$.
Percentage remaining = $0.25 	imes 100\% = 25\%$.

$A$ radioactive isotope has a half-life of $10 \, \text{years}$. What percentage of the original amount of it remains after $20 \, \text{years}$?

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