Question:medium

Half-life period of a radioactive substance is \(10\,\text{min}\). Then amount of substance decayed in \(40\,\text{min}\) will be:

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After \(n\) half-lives: remaining \(=\left(\frac{1}{2}\right)^n\), decayed \(=1-\left(\frac{1}{2}\right)^n\).
Updated On: Jun 16, 2026
  • \(25\%\)
  • \(50\%\)
  • \(75\%\)
  • None of these
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The Correct Option is D

Solution and Explanation

To determine the amount of a radioactive substance that has decayed in a given time, we need to understand the concept of "half-life." The half-life of a radioactive substance is the time required for half of the sample to decay.

In this case, the half-life of the substance is \(10\,\text{min}\). We are asked to find the amount of substance decayed in \(40\,\text{min}\).

First, calculate how many half-lives have passed in \(40\,\text{min}\):

\(\text{Number of half-lives} = \frac{40\,\text{min}}{10\,\text{min}} = 4\)

After each half-life, the amount of remaining substance is halved. Initially, we start with the full amount \(N_0\). After one half-life, the remaining amount is \(\frac{N_0}{2}\), after two half-lives it's \(\frac{N_0}{4}\), and so on.

Following this sequence for four half-lives:

  • After 1st half-life: \(\frac{N_0}{2}\)
  • After 2nd half-life: \(\frac{N_0}{4}\)
  • After 3rd half-life: \(\frac{N_0}{8}\)
  • After 4th half-life: \(\frac{N_0}{16}\)

So, after 4 half-lives, the remaining substance is \(\frac{N_0}{16}\).

The amount that has decayed is:

\(\text{Decayed amount} = N_0 - \frac{N_0}{16} = N_0 \left(1 - \frac{1}{16}\right) = N_0 \cdot \frac{15}{16}\)

The percentage of the decayed substance can be calculated as:

\(\text{Decayed percentage} = \left(\frac{15}{16}\right) \times 100\% = 93.75\%\)

Thus, the percentage of substance decayed in \(40\,\text{min}\) is \(93.75\%\).

Therefore, the correct answer is "None of these" as \(93.75\%\) does not match any of the given options.

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