Question

What is the half life of a radioactive material if $\dfrac{1}{16}$ part of that material is left after one hour?
A. 45 min
B. 30 min
C. 20 min
D. 15 min

Answer 1

Hint: Every radio-active reaction follows first order kinetics, which means that the rate of decay of the element depends is proportional to the first power of the concentration of that element. This states that the element will decay exponentially with the time. Also half-life of a sample of element is known as the time in which the sample will decay to half of its initial concentration.

Formula used:
$t_{1/2} = \dfrac{ln2}{\lambda}$,
$N = N_o e^{-\lambda t}$.

Complete answer:
As the concentration of an element undergoing first order reaction varies as $N = N_o e^{-\lambda t}$ (here, N is the amount decayed after time ‘t’ from a sample of $N_o$ and $\lambda$ is a constant, called decay constant), its given that we have the final concentration as 1/16 of the initial concentration, thus if initial concentration is $N_o$, then $N = \dfrac{1}{16} N_o$. Hence using:
$N = N_o e^{-\lambda t}$
Or $\dfrac{1}{16}N_o = N_o e^{-\lambda t}$
$\implies e^{\lambda t} = 16$
Taking natural log both sides, we get;
$\lambda t = ln 16$
Given, time ‘t’ for this decay is 1 hour or 60 min.
Thus, $\lambda = \dfrac{ln16}{60} min^{-1}$
Now, using formula of half-life, we get;
$t_{1/2} = \dfrac{ln2}{\lambda}$
Also $\lambda = \dfrac{ln16}{60} = \dfrac{4ln2}{60}min^{-1}$ [as $ln(a^b) = b\ ln (a)$]
Hence, $t_{1/2} = \dfrac{ln2}{4ln2/60} = \dfrac{60}{4} =15 min$

So, the correct answer is “Option D”.

Note:
Every reaction, whether nuclear or chemical obeys a certain order of reaction. Reaction can be of zeroth order, first order, second order or higher order. The order is decided as per the behavior of each reactant. If the rate of change of concentration of reactant depends upon the power 2 of the concentration, it’s called second order. In that way, the zeroth order reaction doesn’t really depend upon the concentration of the reactant. When-ever we have a nuclear reaction, we have to assume it first order unless stated.