A radioactive element gets spilled over the floor of a room. Its half-life period is 30 days. If the initial activity is ten times the permissible value, after how many days will it be safe to enter the room?

Radioactive decay follows first-order kinetics. The activity $$A$$ at any time $$t$$ is related to the initial activity $$A_0$$ by

$$A = A_0 \left(\frac{1}{2}\right)^{\,\frac{t}{t_{1/2}}}$$

where $$t_{1/2}$$ is the half-life.

Given data:
  • Half-life $$t_{1/2}=30$$ days
  • Initial activity $$A_0 = 10A_{\text{perm}}$$ (ten times the permissible activity)
  • Required final activity $$A = A_{\text{perm}}$$

Set up the decay equation:

$$A_{\text{perm}} = 10A_{\text{perm}}\left(\frac{1}{2}\right)^{\,\frac{t}{30}}$$

Divide both sides by $$A_{\text{perm}}$$ to cancel it:

$$1 = 10\left(\frac{1}{2}\right)^{\,\frac{t}{30}}$$

Rearrange:

$$\left(\frac{1}{2}\right)^{\,\frac{t}{30}} = \frac{1}{10}$$

Take logarithms (base 10 or natural, any base works) on both sides:

$$\frac{t}{30}\,\log\!\left(\frac{1}{2}\right) = \log\!\left(\frac{1}{10}\right)$$

Solve for $$t$$:

$$t = 30 \times \frac{\log\!\left(\frac{1}{10}\right)}{\log\!\left(\frac{1}{2}\right)}$$

Because $$\log\!\left(\frac{1}{10}\right) = -1$$ (in base-10) and $$\log\!\left(\frac{1}{2}\right) = -0.3010$$,

$$t = 30 \times \frac{-1}{-0.3010} \approx 30 \times 3.322 \approx 99.7\ \text{days}$$

Rounding to the nearest whole day, the safe time is approximately $$100$$ days.

Option D which is: 100 days