What happens to a radioactive material's atom when it disintegrates? Suppose you initial  had radioactive $2^n$ atoms (where $n$ is an integer). Now after a number of halflives the number of left out atoms becomes 1. Now what will happen to it will it disintegrate and the leftover would be half an atom? Now if the reaction stops then the statement "The decaying radioactive atom would never end" then it'll be wrong. 
 A: Radioactive decay is a stochastic process. This means that there is random chance involved, so the exponential model used to represent radioactive does not say exactly how many atoms of the original substance will be left at a given time, rather it tells you the expected value of atoms remaining. If you begin with n=1 atom, after some time the exponential model gives you n=0.5. This does not mean there are 0.5 atoms remaining, it rather means that there is a 0.5 chance that the atom has not decayed yet.
A: Let us suppose that you are indeed left with one unstable nucleus with a half life of $\tau$.
The half life is the time interval during which the probability that the nucleus will decay is $\frac 12$.  
So you start the clock at time $t=0$ and wait for one half life when the time is $t=\tau$.
The probability that the nucleus will decay in that time is $\frac 12$ and the probability that the nucleus will not decay is $1-\frac 12 = \frac 12$.  
So they deacy could happen between a time $t=\tau$ and $t=2\tau$.
Again in that interval of time the probability of a decay is the same as that of not decaying, $\frac 12$.  
The probability of the nucleus not decaying between time $t=0$ and $t=\tau$ and then decaying between $t=\tau$ and $t=2\tau$ is $\frac 12 \times \frac 12 =\frac 14 =\frac {1}{2^2}$
The probability of the nucleus not decaying between time $t=0$ and $t=2\tau$ and then decaying between $t=2\tau$ and $t=3\tau$ is $\frac 14 \times \frac 12 =\frac 18=\frac {1}{2^3}$ 
$...$ 
The probability of the nucleus not decaying between time $t=0$ and $t=n\tau$ and then decaying between $t=n\tau$ and $t=(n+1)\tau$ is $\frac {1}{2^n} \times \frac 12 =\frac {1}{2^{(n+1)}}$ 
When one adds up all the probabilities of decaying between $t=0$ and $t=\tau$; $t=\tau$ and $t=2\tau$; $t=2\tau$ and $t=3\tau$ etc which is $$\frac 12 + \frac 14+ \frac 18 + .....=1$$
as expected ie if you wait an infinite time the nucleus will have decayed at some time.
So in the need you may have to wait a very long time for the one remaining nucleus to decay and the probability of decay in the first half life is $\frac 12$.  
The exponential decay function is only a ((very) good) approximation if you are dealing with large numbers of nuclei and such that the statistical fluctuations in the rate of decay are very small compared with the rate of decay.  

The Phet Alpha Decay simulation is worth a look at?
A: Radioactivity does not mean that an atom disappears. It means that the atom splits into one or more different smaller atoms or fragments of atoms. Very little mass is lost. The mass of all the fragments is not much less than the mass of the original atoms. 
When the last radioactive atom has decayed, the process of radioactive disintegration stops. It does not go on forever, as the mathematical model suggests. You cannot have fractions of an atom left over, and fractions of an atom cannot decay at any time. Like the average Western family being 2.4 children, there are no families with 0.4 of a child.
If the fragments are unstable they will also decay, with a different half-life. 
