Question

A radioactive substance has a half life of four months. Three fourth of the substance will decay in:
A. Three months
B. Four months
C. Eight months
D. Twelve months

Answer 1

Hint:Half-life of radioactive decay is the interval of time required for one half of the atomic nuclei of a radioactive sample to decay, use this to find out the decay constant. To find out the time required for three fourth of the substance to decay, we need to find out how much substance will be left after three fourth of it has decayed. Then equate these values in the decay law to find out the value for time taken.
Formula used:
$N = {N_o}{e^{ - \lambda t}}$
Where: ${N_o} = $ Initial concentration, $N = $ Concentration after time $t$
${t_{\dfrac{1}{2}}} = \dfrac{{0.693}}{\lambda }$
: ${t_{\dfrac{1}{2}}} = $ Half Life , $\lambda = $ decay constant

Complete step by step answer:
Radioactive elements are made up of atoms whose nuclei are unstable and give off atomic radiation as a part of a process of attain stability. The emission of radiation transforms radiation transforms radioactive atoms into another chemical element which may be stable or may be radioactive such that it undergoes further decay.
There are 28 naturally occurring radioactive, consisting of 34 radionuclides. These 34 are known as primordial nuclides.
Long lived radioactive elements such as uranium, thorium, and potassium and any of the decay products, such as radium and radon are examples of NORM. These elements have always been present in the earth’s crust and atmosphere.
The time interval required for one half of the atomic nuclei of a radioactive sample to decay.
The most common type of radiation is alpha, gamma radiation .Radioactivity decay rates are normally stated in terms of their half-lives. The different types of radioactivity lead to different decay paths.
Decay constant: it is the probability of decay per unit time. The number of parent nuclides $P$ therefore decreases with time $t$ as $\dfrac{{dP}}{P}dt = - \lambda $ . The decay constant is related to the half-life of the nuclide ${T_{1/2}}$ through ${t_{1/2}} = \ln 2/\lambda $ .
Half-life, in radioactivity, the interval of time required for one half of the atomic nuclei of a radioactive sample to decay.
The half-life of a first order reaction is a constant that is related to the rate constant for the reaction:
${t_{\dfrac{1}{2}}} = \dfrac{{0.693}}{\lambda }$
Where: ${t_{\dfrac{1}{2}}} = half\,life,\lambda = decay\,constant$
Now, comes to the solution;
$\lambda = \dfrac{{0.693}}{4} \Rightarrow 0.173$
Let us assume the initial concentration will be ${N_o}$
It is given that the half-life of given substances is 4 months.
The amount of substance left after 4 months $\dfrac{1}{2}{N_o}$
The amount of substance left after $t$ months ${N_o} - \dfrac{3}{4}{N_o}$ $ = \dfrac{1}{4}{N_o}$
We know that:
$N = {N_o} \times {e^{ - \lambda \times t}}$
Where: ${N_o} = $ initial concentration, $N =$ concentration after time
Substituting the values we got above:
$\dfrac{1}{4}{N_o} = {N_o}{e^{ - 0.173 \times t}}$
Solving it further:
$\dfrac{1}{4} = {e^{ - 0.173 \times t}}$
Using the property for Natural logarithmic function, we get:
${\log _e}\dfrac{1}{4} = - 0.173 \times t$
Hence, $t = 8.11 \sim 8months$
That is three fourth of the radioactive sample will decay in 8 months.
Hence, option C is the correct answer.

Note:
Alternate method to solve this question could be,
The amount of substance left after 4 months $\dfrac{1}{2}{N_o}$
The amount of substance left after $x$ months ${N_o} - \dfrac{3}{4}{N_o}$ $ = \dfrac{1}{4}{N_o}$
$\therefore $ Since $4$ months is required to reduce the sample to $\dfrac{1}{2}{N_o}$ .
Hence, $x$ months will be required to convert the sample to $\dfrac{1}{4}{N_o}$
$\dfrac{1}{4}{N_o}$ can also be written as $\dfrac{1}{{{2^2}}}{N_o}$
Which means the concentration is reduced by half.
We know that:
$T = n \times {t_{\dfrac{1}{2}}}$
Where: $T = time\,taken\,for\,concentration\,to\,reduce\,to\,given\,value$
We know, the value of $n = 2,{t_{\dfrac{1}{2}}} = 4months$
Substituting these values we get,
$T = 2 \times 4 \Rightarrow 8months$