Question

Gold$$-198$$ has a half-life of $$2.7\text{ days}$$.How much of a $$96g$$ sample of Gold$$-198$$ will be after $$8.1$$ days?

Answer 1

Hint: Radioactive decay follows first order kinetics.
Use the concept of half-life of first order kinetics and the formula of half-life for first order kinetics is
$$t_{{\displaystyle \frac{1}{2}}}={\displaystyle \frac{0.693}{\lambda }}$$ Here $$\lambda$$ is rate constant of gold
And use the equation of first order to find the remaining concentration after given time

Complete step-by-step answer:
 Now it is given in the question that Gold$$-198$$ has a half-life of$$2.7\text{ days}$$.
That means the value of$$t_{{\displaystyle \frac{1}{2}}}=2.7\text{ days}$$.
Now using the equation of half-life which is $$t_{{\displaystyle \frac{1}{2}}}={\displaystyle \frac{0.693}{\lambda }}$$ we will find the value of rate constant$$\lambda$$.
The value of $$\lambda$$ will be $$\lambda ={\displaystyle \frac{0.693}{t_{{\displaystyle \frac{1}{2}}}}}$$
Putting the value of half time of gold we will get the rate constant which is
$$\lambda ={\displaystyle \frac{0.693}{2.7}}$$
$$\lambda =0.256\text{ day}{\text{s}}^{-1}$$
Now the equation of first order kinetics is $$N=N_0\times e^{-\lambda t}$$
Here, $$N$$ is the amount left after $$t$$ time
$$N_0$$ Is the initial amount which is = $$96g$$
$$\lambda$$ Is the rate constant whose value we had founded earlier = $$0.26$$
And the value of $$t$$ is given in question which is $$t=8.1\text{ days}$$
Now putting all the values in the equation $$N=N_0\times e^{-\lambda t}$$ we will get
$$N=96\times e^{-0.256day{s}^{-1}\times 8.1days}$$
 Now solving it we will get $$N=12g$$

Thus the amount of Gold$$-198$$ left after $$8.1\text{ days}$$ is $$12grams$$

Note: The values unit should be the same at all the places and the thing which is to be noted that all the radioactive decays follow the first-order kinetics. Half-life, in radioactivity, the interval of time required for one-half of the atomic nuclei of a radioactive sample to decay (change spontaneously into other nuclear species by emitting particles and energy), or, equivalently, the time interval required for the number of disintegrations per second of radioactive material to decrease by one-half.