Solved: PHYSICS Question for NEET | Sushrut

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NEET PHYSICSEasy

In a radioactive substance at t = 0, the number of atoms is $8 \times 10^4$ . Its half-life period is 3 yr. The number of atoms equal to $1 \times 10^4$ will remain after an interval of:

A

9 yr

B

8 yr

C

6 yr

D

24 yr

Step-by-Step Solution

Identify the Principle: Radioactive decay follows first-order kinetics. The number of undecayed nuclei $N$ remaining after $n$ half-lives is given by the formula $N = N_0 (\frac{1}{2})^n$ , where $N_0$ is the initial number of nuclei .
Analyze Given Data:

Initial number of atoms ( $N_0$ ) = $8 \times 10^4$ .
Final number of atoms ( $N$ ) = $1 \times 10^4$ .
Half-life period ( $T_{1/2}$ ) = 3 years.

Calculate Number of Half-lives ( $n$ ):

$\frac{N}{N_0} = \frac{1 \times 10^4}{8 \times 10^4} = \frac{1}{8}$ .
We know that $\frac{1}{8} = (\frac{1}{2})^3$ .
Therefore, the number of half-lives passed, $n = 3$ .

Calculate Total Time ( $t$ ):

Total time $t = n \times T_{1/2}$ .
$t = 3 \times 3 \text{ years} = 9 \text{ years}$ .

Conclusion: The atoms will reduce to the specified amount after 9 years .

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