You must have noticed that half-life is defined for various radioactive elements and other substances too.Why is it that always the half-life is defined?Why not a quarter-life or full life?
[Physics] Half-life of substances
half-lifenuclear-physicsradioactivity
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The right way to think about this is that, over 5,730 years, each single carbon-14 atom has a 50% chance of decaying. Since a typical sample has a huge number of atoms1, and since they decay more or less independently2, we can statistically say, with a very high accuracy, that after 5,730 years half of all the original carbon-14 atoms will have decayed, while the rest still remain.
To answer your next natural question, no, this does not mean that the remaining carbon-14 atoms would be "just about to decay". Generally speaking, atomic nuclei do not have a memory3: as long as it has not decayed, a carbon-14 nucleus created yesterday is exactly identical to one created a year ago or 10,000 years ago or even a million years ago. All those nuclei, if they're still around today, have the same 50% probability of decaying within the next 5,730 years.
If you like, you could imagine each carbon-14 nucleus repeatedly tossing a very biased imaginary coin very fast (faster than we could possibly measure): on each toss, with a very, very tiny chance, the coin comes up heads and the nucleus decays; otherwise, it comes up tails, and the nucleus stays together for now. Over a period of, say, a second or a day, the odds of any of the coin tosses coming up heads are still tiny — but, over 5,730 years, the many, many tiny odds gradually add up to a cumulative decay probability of about 50%.
1 A gram of carbon contains about 0.08 moles, or about 5 × 1022 atoms. In a typical natural sample, about one in a trillion (1 / 1012) of these will be carbon-14, giving us about 50 billion (5 × 1010) carbon-14 atoms in each gram of carbon.
2 Induced radioactive decay does occur, most notably in fission chain reactions. Carbon-14, however, undergoes spontaneous β− decay, whose rate is not normally affected by external influences to any significant degree.
3 Nuclear isomers and other excited nuclear states do exist, so it's not quite right to say that all nuclei of a given isotope are always identical. Still, even these can, in practice, be effectively modeled as discrete states, with spontaneous transitions between different states occurring randomly with a fixed rate over time, just as nuclear decay events do.
The half life is not a guarantee of faster rate of energy release: just faster rate of disintegration. You need to multiply that by the energy released per disintegration to get the energy release per second.
All other things being equal (same atomic mass, same energy released), you need to bring less material (mass) along if you try to generate a certain amount of energy, if that material has a shorter half life.
The simple way to see that: if the total energy needed is released by 3000 atoms disintegrating, and the journey takes one half life, then I need to bring 6000 atoms along (half disintegrate). But if the half life is twice as short, I only need to bring 4000 atoms: after half the flight 2000 disintegrated, and another 1000 in the second half of the flight.
That demonstrates another problem: if you need constant power from your source, you need a longer half life...
In other words - it depends.
Best Answer
The half-life $t_{1/2}$ is defined so that the amount of stuff you have $N(t)$ at some time $t$ is $$ N(t) = N_0 \cdot 2^{-t/t_{1/2}} $$ where $N_0$ is how much stuff you start with and you're familiar with the constant $2$.
This definition has several advantages. First, it's easy to explain, since scientists of all ages are familiar with the constant $2$. Second, it's unambiguous. If you asked about a "quarter-life," do you mean when a quarter of your original material remains $t_{1/4}$, which is after two half-lives, or when a quarter of your original material has decayed? The second option is not $\frac12 t_{1/2}$, because the decay is exponential and not linear; it's $t_{3/4} = t_{1/2}\log_2\frac43$, ugh.
"Full life" is not an option because the number of remaining decay-ers approaches zero only asymptotically.
When you start to do calculus, the same sort $\log_2$ ugliness comes up again. Another, probably nicer, way to write the decay equation is $$ N(t) = N_0 e^{-t/\tau} $$ where $e = \frac1{0!} + \frac1{1!} + \frac1{2!} + \cdots \approx 2.7183$ is the base of the natural logarithms. The constant $\tau$ is referred to by several names, depending on your mood and the experience level of the person you're communicating with: sometimes $\lambda \equiv 1/\tau$ is called the "decay constant," and sometimes $\tau$ itself is referred to as the "$e$-folding time" or simply "the lifetime."
A nice feature of the exponential form is that, if you assume that every decay is detectable (or that a known fraction of the decays are detectable) then you can find the measured activity of your source: $$ A(t) = -\frac{dN}{dt} = \frac{N_0}{\tau}e^{-t/\tau} $$ This means that if you can measure what your source's activity is, and measure long enough to watch it change, you have effectively weighed the radioactive part of the source by fixing $N_0$, even if $N_0$ may be only a few millions of atoms.