Question

At a given instant there are 50% undecayed radioactive nuclei in a sample. After 15 seconds the number of undecayed nuclei reduced to 25%. The time in which the number of undecayed nuclei will further reduce to 6.25% of the reduced number of nuclei is x seconds. Find the value of $\frac{x}{10}$.

Answer 1

6

Answer 2

1. Determine the half-life ($T_{1/2}$):

At a given instant, there are 50% undecayed nuclei. After 15 seconds, the number of undecayed nuclei reduces to 25%.
This means the amount of undecayed nuclei has halved (from 50% to 25%).
Therefore, the time taken for one half-life is 15 seconds.
$T_{1/2} = 15 \text{ s}$.

2. Interpret the target percentage:

The question asks for the time 'x' in which the number of undecayed nuclei will further reduce to 6.25% of the reduced number of nuclei.
The "reduced number of nuclei" refers to the amount present after 15 seconds, which is 25% of the initial amount ($N_0$).
So, the starting point for this further reduction is $N_{start} = 0.25 N_0$.
The target amount for the undecayed nuclei is 6.25% of $N_{start}$:
$N_{final} = 0.0625 \times N_{start} = 0.0625 \times (0.25 N_0)$.
$N_{final} = 0.015625 N_0$.

3. Calculate the number of half-lives (n) for this reduction:

We use the formula for radioactive decay: $N = N_{start} \left(\frac{1}{2}\right)^n$.
Substitute the values:
$0.015625 N_0 = 0.25 N_0 \left(\frac{1}{2}\right)^n$.
Divide both sides by $0.25 N_0$:
$\frac{0.015625}{0.25} = \left(\frac{1}{2}\right)^n$.
$0.0625 = \left(\frac{1}{2}\right)^n$.
Convert 0.0625 to a fraction: $0.0625 = \frac{625}{10000} = \frac{1}{16}$.
So, $\frac{1}{16} = \left(\frac{1}{2}\right)^n$.
Since $\frac{1}{16} = \left(\frac{1}{2}\right)^4$, we have:
$n = 4$.
This means 4 half-lives are required for this reduction.

4. Calculate the time 'x':

The time 'x' is the duration for these 4 half-lives:
$x = n \times T_{1/2} = 4 \times 15 \text{ s} = 60 \text{ s}$.

5. Find the value of $\frac{x}{10}$:

$\frac{x}{10} = \frac{60}{10} = 6$.