Question

What is radioactivity? State the law of radioactivity decay. Show that radioactive decay is exponential in nature.
The half-life of radium is 1600 years. How much time does $$1g$$ of radium take to reduce to $$0.125g$$?

Answer 1

Hint: Radioactivity is a process of spontaneous disintegration or breaking of unstable atomic nucleus by the emission of radiation like $\alpha -\text{ray}$ or $\beta -\text{ray}$or $\lambda -\text{ray}$. The rate of disintegration will be directly proportional to the amount of substance present at that instant.
For the given problem first calculate the decay constant for the given case. And from that calculate the time required to decay from decay law.

Formula Used:
Half life, $T={\displaystyle \frac{0.693}{\lambda }}$
$N=N_0{e}^{-\lambda t}$
$\begin{array}{rl}& \log a-\log b=\log \left({\displaystyle \frac{a}{b}}\right)\\ & \log \left(ab\right)=\log a+\log b\\ & \log a^b=b\log a\\ & {\log }_{a}a=1\end{array}$

Complete step-by-step answer:
The phenomenon by virtue of which substance, spontaneously, disintegrates by emitting certain radiations is called radioactivity.
Laws of Radioactive-decay: Followings are the laws of radioactive decay.
(i) Radioactivity is due to the disintegration of a nucleus.
(ii) Rate of disintegration is not affected by external conditions like temperature and pressure etc. This observation leads us to the conclusion that the seat of radioactivity must be deep inside the atom, i.e., nucleus. Orbital electrons have nothing to do with the phenomenon of radioactivity.
(iii) Law of conservation of charge holds good in radioactivity. It means the total charge before disintegration and that after disintegration must be the same.
(iv) The disintegration is accompanied by the emission of energy in the form of $\alpha$ ,$\beta$ and $\gamma$ rays either single or all at a time. Emission of these radioactive radiations bring about following changes:
(a) Emission of $\alpha -\text{particle}$ results in a decrease in its atomic number by $2$ and a decrease in its atomic weight by $4$.
(b) Emission of $\beta -\text{particle}$ results in an increase in atomic number by one while its atomic weight remains unaffected.
(c) Emission of $\gamma -\text{rays}$ results neither in a change of atomic number nor in change of atomic weight.
(v) Each product of disintegration is a new element having physical and chemical properties different from those of the parent atom.
(vi) Rate of disintegration of the radioactive substance, at any instant, is directly proportional to the number of atoms present at that instant. This is known as the statistical law of radioactivity.
Let $N$ be the number of atoms of radioactive samples at any instant. The rate of disintegration will be $-{\displaystyle \frac{dN}{dt}}$ (negative sigh because the number of atoms is decreasing with time). So according to law of radioactive disintegration $-{\displaystyle \frac{dN}{dt}}\alpha N,\text{ or }{\displaystyle \frac{dN}{dt}}=-\lambda N$
$\lambda =\text{ radioactive decay constant and depends upon nature of substance}$
Rewriting ${\displaystyle \frac{dN}{N}}=-\lambda dt$
 integrating both sides.
$\begin{array}{rl}& \int {\displaystyle \frac{dN}{N}}=\int -\lambda dt+C\\ & \Rightarrow {\log }_{e}N=-\lambda t+C\end{array}$
C is a constant of integration that can be determined from initial condition.
At,
$\begin{array}{rl}& t=0,N=N_0\\ & \Rightarrow {\log }_e{N}_0=0+C\\ & \Rightarrow C={\log }_e{N}_0\end{array}$
So,
 $\begin{array}{rl}& {\log }_{e}N=-\lambda t+{\log }_e{N}_0\\ & \Rightarrow {\log }_{e}N-\log N_0=-\lambda t\end{array}$
$\Rightarrow {\log }_e\left[{\displaystyle \frac{N}{N_0}}\right]=-\lambda t$
Or, $N=N_0{e}^{-\lambda t}$
This gives the number of radioactive substances left at any time. Which is decreasing exponentially.
Half life of a radioactive substance is defined as the time during which the number of atoms of the substance are reduced to half their original value.
So at half life i.e.at $t=t_{{\displaystyle \frac{1}{2}}}=T,N={\displaystyle \frac{N_0}{2}}$
Now above equation becomes
${\displaystyle \frac{N_0}{2}}=N_0{e}^{-\lambda T}\text{ or , }2=e^{\lambda T}$
Taking natural logarithm.
$$\begin{array}{rl}& {\log }_{e}2={\log }_e{e}^{\lambda T}=\lambda T{\log }_{e}e\\ & \Rightarrow 2.303{\log }_{10}2=\lambda T\\ & \Rightarrow T={\displaystyle \frac{1}{\lambda }}\times 2.303{\log }_{10}2\\ & \Rightarrow T={\displaystyle \frac{0.693}{\lambda }}\end{array}$$

In the given problem the half-life of radium is 1600 years.
So $T=1600\text{years}$
From above we get $$T={\displaystyle \frac{0.693}{\lambda }}$$
So $\lambda ={\displaystyle \frac{0.693}{1600}}$
Initial amount of radium is $$1g$$.i.e. $N_0=1g$
Final amount i.e.$N=0.025g$
According to decay law $N=N_0{e}^{-\lambda t}$
Taking logarithm
$$\begin{array}{rl}& {\log }_{e}N={\log }_e\left(N_0{e}^{-\lambda t}\right)\\ & \Rightarrow {\log }_{e}N={\log }_e{N}_0+{\log }_e{e}^{-\lambda t}\\ & \Rightarrow {\log }_{e}N-{\log }_e{N}_0=-\lambda t{\log }_{e}e\\ & \Rightarrow -\lambda t={\log }_{e}N-{\log }_e{N}_0\\ & \Rightarrow t={\displaystyle \frac{{\log }_e\left({\displaystyle \frac{N}{N_0}}\right)}{-\lambda }}={\displaystyle \frac{{\log }_e\left({\displaystyle \frac{0.025}{1}}\right)}{-{\displaystyle \frac{0.693}{1600}}}}\\ & \Rightarrow t={\displaystyle \frac{-1.60}{-{\displaystyle \frac{0.693}{1600}}}}(\because {\log }_{e}0.025=-1.60)\\ & \Rightarrow t=3699\text{years}\end{array}$$
So after $$3699\text{years}$$The amount of radium will be $0.025g$

Note: During the calculation like this the logarithm value is most important. If you take natural logarithm the value will differ from common logarithm. Also note that the life time of a radioactive substance is infinity i.e. a radioactive substance will emit radiation for eternity.
Also note that if $T$is the half life of a sample then the number of nuclei left undecayed after${}^{\prime }{n}^{\prime }$half life is given by the relation.
${\displaystyle \frac{N}{N_0}}={\left({\displaystyle \frac{1}{2}}\right)}^n$