Question

A substance is kept for two hours and three-fourth of that substance disintegrates during this period. The half-life of the substance is:
(A) 2hrs
(B) 1 hr
(C) 30min
(D) 4hrs

Answer 1

Hint : The half-life of a substance is the time required for a substance to reduce to its half the amount of its initial amount. In the above question time for three-fourth completion of the reaction is given. We will apply the formula of half-life and equate it with the given time.
$t{}^{1}/{}_{2}=\dfrac{1}{2}t{}^{3}/{}_{4}$

Complete step by step solution :
Let’s, see what the solution is
Case 1: when the substance reduces to three-fourth of its initial value
Let us consider ‘a’ to be the initial amount of the substance.
Amount of the substance left after disintegration $=a-\dfrac{3}{4}a=\dfrac{1}{4}a$
Let ‘k’ be the disintegration constant
We know the formula for calculating the disintegration constant is:
$k=\dfrac{2.303}{t{}^{3}/{}_{4}}\log \dfrac{a}{a-\dfrac{3}{4}a}$
Now, $k=\dfrac{2.303}{t{}^{3}/{}_{4}}\log \dfrac{a}{\dfrac{1}{4}a}$
$k=\dfrac{2.303}{t{}^{3}/{}_{4}}\log 4$
$k=\dfrac{2.303\times 0.6020}{t{}^{3}/{}_{4}}$
$k=\dfrac{1.3864}{t{}^{3}/{}_{4}}$…….equation 1

Case2: when the substance reduces to half of its initial value
The formula for half-life is
$t{}^{1}/{}_{2}=\dfrac{0.6932}{k}$……..equation 2
Now, the disintegration constant is same for both the cases
Therefore, on transformation and equating both the equation 1 and equation 2
We get,
$t{}^{1}/{}_{2}=\dfrac{0.6932}{1.3864}t{}^{3}/{}_{4}$
$t{}^{1}/{}_{2}=\dfrac{1}{2}t{}^{3}/{}_{4}$
Where, $t{}^{1}/{}_{2}$ is the half-life and $t{}^{3}/{}_{4}$ is the time period for the three-fourth completion of the reaction
On putting value of $t{}^{3}/{}_{4}$= 2, given in the question
$t{}^{1}/{}_{2}=1$
Hence, the answer for the given question is 1hr

So, the correct answer is “Option B”.

Note : The notion of half-life is used to know when atoms will undergo radioactive decay. It is used to know about exponential decay or non-exponential decay.