Question

Radioactive element decays to form a stable nuclide, then the rate of decay of reactant is shown by

Answer 1

2

Answer 2

The rate of decay of a radioactive element is given by the formula $R(t) = -\frac{dN}{dt} = \lambda N(t)$, where N(t) is the number of radioactive nuclei at time t and $\lambda$ is the decay constant.

For a first-order radioactive decay, the number of radioactive nuclei at time t is given by $N(t) = N_0 e^{-\lambda t}$, where $N_0$ is the initial number of radioactive nuclei at t=0.

Substituting this into the rate equation, we get $R(t) = \lambda N_0 e^{-\lambda t}$.

This equation shows that the rate of decay decreases exponentially with time.

At t=0, the rate of decay is $R(0) = \lambda N_0$, which is the maximum rate.

As time t increases, $e^{-\lambda t}$ decreases, and hence $R(t)$ decreases.

As $t \to \infty$, $e^{-\lambda t} \to 0$, and hence $R(t) \to 0$.

The graph of $R(t)$ versus t is an exponential decay curve, starting from a positive value at t=0 and asymptotically approaching zero as t approaches infinity.

Now let's examine the given graphs. The first graph shows a curve that starts at 0, increases to a maximum, and then decreases to 0. This is not an exponential decay curve.

The second graph shows a curve that starts at a maximum value on the y-axis at t=0 and decreases with time, approaching the x-axis as t increases. This is consistent with an exponential decay curve.

Therefore, the second graph represents the rate of decay of the reactant as a function of time.