The isotope \(_{19}K^{42}\) has a half-life of about 12 hours. What fraction of the initial concentration of \(_{19}K^{42}\) remains after 48 hours?

\(\frac{1}{16}\)

The correct answer is option (4) \(\frac{1}{16}\).

The concept of half-life is used to describe the time it takes for half of a substance to decay or transform. The formula for calculating the remaining fraction after a certain number of half-lives is given by:

\(\text{Remaining fraction} = \left( \frac{1}{2} \right)^{\text{Number of half-lives}} \)

In this case, the isotope \( _{19}K^{42} \) has a half-life of 12 hours. To find the fraction remaining after 48 hours, we need to determine how many half-lives have passed in that time period. We can do this by dividing the total time (48 hours) by the half-life (12 hours):

\(\text{Number of half-lives} = \frac{\text{Total time}}{\text{Half-life}} = \frac{48 \, \text{hours}}{12 \, \text{hours/half-life}} = 4 \, \text{half-lives}\)

Now, we can use the formula to find the remaining fraction:

\(\text{Remaining fraction} = \left( \frac{1}{2} \right)^4 \)

\(\text{Remaining fraction} = \frac{1}{2^4} \)

\(\text{Remaining fraction} = \frac{1}{16} \)

So, after 48 hours, only \(\frac{1}{16}\) of the initial concentration of \( _{19}K^{42} \) remains. This means that over the course of four half-lives, the original amount has undergone multiple cycles of halving, leading to a significant reduction in quantity.