The half-life of radioisotopic bromine-82 is $$36$$ hours. The fraction which remains after one day is ______ $$\times 10^{-2}$$.
(Given antilog $$0.2006 = 1.587$$)

We need to find the fraction of radioisotopic bromine-82 remaining after one day (24 hours), given its half-life is 36 hours.

The formula for radioactive decay is $$\frac{N}{N_0} = \left(\frac{1}{2}\right)^{t/t_{1/2}},$$ where $$t_{1/2} = 36\text{ hours}$$ and $$t = 24\text{ hours}.$$ Hence, $$\frac{t}{t_{1/2}} = \frac{24}{36} = \frac{2}{3},$$ so $$\frac{N}{N_0} = \left(\frac{1}{2}\right)^{2/3}.$$

Taking logarithm (base 10) of this expression gives

$$\log\left(\frac{N}{N_0}\right) = \frac{2}{3}\times\log\left(\frac{1}{2}\right) = \frac{2}{3}\times(-0.3010) = -0.2007.$$

Thus,

$$\frac{N}{N_0} = \text{antilog}(-0.2007) = \frac{1}{\text{antilog}(0.2007)}$$

and since antilog(0.2006) = 1.587,

$$\frac{N}{N_0} = \frac{1}{1.587} = 0.6301 \approx 63\times10^{-2}.$$

Therefore, the fraction remaining is 63 × 10−2.