Question

$1.0 , \text{g}$ of a radioactive isotope was found to reduce to $125 , \text{mg}$ after 24 hours. The half-life of the isotope is:

A) 8 hours
B) 24 hours
C) 6 hours
D) 4 hours

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Answer

The correct answer is A.

First, using the decay formula for radioactive isotopes: $$ N = \left(\frac{1}{2}\right)^{n} \times N_{0} $$

Here, initially we have: $$ N_{0} = 1000 , \text{mg} $$ and after 24 hours it reduces to: $$ N = 125 , \text{mg} $$

So, $$ 125 , \text{mg} = \left(\frac{1}{2}\right)^{n} \times 1000 , \text{mg} $$

Next, we solve for n: $$ \left(\frac{1}{2}\right)^{n} = \frac{125}{1000} = \frac{1}{8} $$

Recognizing that: $$ \left(\frac{1}{2}\right)^{n} = \left(\frac{1}{2}\right)^{3} $$

Thus, $$ n = 3 $$

The number of half-lives ($ t_{1/2} $) is 3. Given the total time is 24 hours, the half-life can be calculated as: $$ \text{Half-life time} = \frac{24 , \text{hours}}{3} = 8 , \text{hours} $$

Therefore, the half-life of the isotope is 8 hours.

Final Answer: A


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