Question

The rate constant (k) for the decay of the radioactive isotope I-131 is 3.6 x 10-3 hours 1. The slope of which of the following graphs is correct for the decay and could be used to confirm the value of k?

Answer 1

The correct slope to find the rate constant (k) for the decay of I-131 is the negative of the decay constant, which would be -3.6 x 10^-3 on a plot of ln(activity) versus time.

Step-by-step explanation:

The rate constant for the decay of the radioactive isotope I-131 is 3.6 x 10-3 hours-1. When plotting a graph for the decay, one would typically use a logarithmic representation of the number of atoms remaining versus time. The slope of the graph for a first-order decay process should be equal to the negative of the decay constant (k).

Therefore, the correct slope to confirm the value of k would be -3.6 x 10-3 on a graph where the y-axis is the natural logarithm (ln) of the number of atoms or activity (A), and the x-axis represents time (t). This aligns with the linear form of the decay equation ln(A) = -kt + ln(A0), where A is the activity, k is the decay constant, t is time, and A0 is the initial activity.

Answer 2

The correct graph to confirm the rate constant would be a plot of ln[A] versus time, where the slope of the line is the negative value of k, which is -3.6 x 10^-3 hours^-1 for I-131.

Step-by-step explanation:

To confirm the value of the rate constant (k) for the decay of the radioactive isotope I-131, which is given as 3.6 x 10-3 hours-1, we would look for a graph that represents first-order decay. For first-order reactions, a plot of the natural logarithm of the remaining quantity of the substance (ln[A]) versus time (t) will yield a straight line with a slope equal to the negative value of the rate constant (-k). Therefore, the correct graph to confirm the rate constant of I-131 decay would show a straight line with a slope of -3.6 x 10-3 hours-1.