Question No. 3
An accident in a nuclear laboratory resulted in deposition of a certain amount of radioactive material of half-life 18 days inside the laboratory. Tests revealed that the radiation was 64 times more than the permissible level required for safe operation of the laboratory. What is the minimum number of days after which the laboratory can be considered safe for use?
(A) 64
(B) 90
(C) 108
(D) 120

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Solution:

Let $N_0$ be the initial amount of radioactive material, and $N$ be the permissible safe level.
Given:

• Initial radiation is 64 times the safe level: $N_0=64N$.

• Half-life $T_{1\mathrm{/}2}=18$ days.

We need to find the time $t$ (in days) such that the amount decays to $N$:

$$N=N_0{\left(\frac{1}{2}\right)}^{t\mathrm{/}{T}_{1\mathrm{/}2}}$$

Substitute $N_0=64N$:

$$N=64N{\left(\frac{1}{2}\right)}^{t\mathrm{/}18}$$

Divide both sides by $N$:

$$1=64{\left(\frac{1}{2}\right)}^{t\mathrm{/}18}$$$${\left(\frac{1}{2}\right)}^{t\mathrm{/}18}=\frac{1}{64}$$

Note that $64=2^6$, so $\frac{1}{64}={\left(\frac{1}{2}\right)}^6$.
Thus:

$${\left(\frac{1}{2}\right)}^{t\mathrm{/}18}={\left(\frac{1}{2}\right)}^6$$

Since the bases are equal:

$$\frac{t}{18}=6$$$$t=6\times 18=108\text{ days}$$

Final Answer:

$$\boxed{{\displaystyle 108}}$$