Physics Q20 – NEET-UG 2023

The half life of a radioactive substance is 20 minutes. In how much time, the activity of substance drops to

{(\frac{1}{16})}^{t h}

of its initial value?

Clue 1 of 3: Radioactive Decay

Radioactive decay is a first-order process where the activity of a substance decreases exponentially over time.

🥷

Ninja Strategy•Half-life Progression

Quickly determine the number of half-lives needed for the activity to reach 1/16 (which is 4 half-lives), then multiply by the given half-life duration.

🧠 Full Solution Path

Step 1: Determine the Number of Half-lives✦ Active

The activity of a radioactive substance decreases by a factor of $\frac{1}{2}$ for every half-life. We are given that the activity drops to $\frac{1}{16}$ of its initial value. We can write this as:

A = A_{0} {(\frac{1}{2})}^{n}

Where $A$ is the final activity, $A_{0}$ is the initial activity, and $n$ is the number of half-lives. Substituting the given ratio:

\frac{1}{16} = {(\frac{1}{2})}^{n}

Since $\frac{1}{16} = {(\frac{1}{2})}^{4}$ , we have:

{(\frac{1}{2})}^{4} = {(\frac{1}{2})}^{n} ⟹ n = 4

💡 Teacher's Secret Hint

Remember that the fraction of remaining activity is always a power of $\frac{1}{2}$ .

Step 2: Calculate the Total Time○ Expand

The total time $t$ required for the activity to drop is the product of the number of half-lives ( $n$ ) and the half-life ( $T_{1 / 2}$ ). We are given $T_{1 / 2} = 20$ minutes and we found $n = 4$ .

t = n \times T_{1 / 2}

Substituting the values:

t = 4 \times 20 minutes = 80 minutes

💡 Teacher's Secret Hint

Ensure units are consistent throughout the calculation.

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Physics Question 20 – NEET-UG 2023

Clue 1 of 3: Radioactive Decay

🧠 Full Solution Path