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Question: The half-life of a radioactive substance is \(6\) hours. The amount of the substance that has disint...

The half-life of a radioactive substance is 66 hours. The amount of the substance that has disintegrated when 36gm36gm of it undergoes decay for 1818 hours is
A. 31.5gm31.5gm
B. 4.5gm4.5gm
C. 18gm18gm
D. 9gm9gm

Explanation

Solution

First use the equation for the rate of radioactive decay, i.e. ln⁡[NN0]=−kt\ln \left[ {\dfrac{N}{{{N_0}}}} \right] = - kt , for the decay of half of the radioactive substance (i.e. N=50N = 50 , N0=100{N_0} = 100 and t=6hours=360min⁡t = 6hours = 360\min ) and calculate the value of kk . Then substitute this value of kk in ln⁡[NN0]=−kt\ln \left[ {\dfrac{N}{{{N_0}}}} \right] = - kt , where t=18hours=1080min⁡t = 18hours = 1080\min and N0=36grams{N_0} = 36grams , to calculate the value of NN . Then calculate the value of N0−N{N_0} - N to reach the solution.

Complete answer:
Before moving on to the mathematical calculations, it would be beneficial for us to discuss radioactive disintegration (radioactive decay) and half-life.
Radioactive disintegration also called radioactive decay is the spontaneous decay of an atomic nucleus which results in the release of matter and energy from the nucleus. A radioactive substance has unstable nuclei that do not have the required binding energy (the energy required to separate all the constituents of the nucleus away from each other) to keep the nucleus intact.
Half-life – If we have a given amount of radioactive substance, then the time taken for half of that substance to decay is known as the half-life of that radioactive substance.
We know that the equation of the rate of radioactive decay is
ln⁡[NN0]=−kt\ln \left[ {\dfrac{N}{{{N_0}}}} \right] = - kt
Here, N=N = The amount of radioactive substance left after time tt
N0={N_0} = The initial amount of radioactive substance
k=k = The rate of constant of radioactive decay
t=t = Time
So, for the amount of the substance to decrease by half, N=50N = 50
N0=100{N_0} = 100
t=6hours=360min⁡t = 6hours = 360\min
So, ln⁡[50100]=−k×360\ln \left[ {\dfrac{{50}}{{100}}} \right] = - k \times 360
ln⁡[12]=−k×360\ln \left[ {\dfrac{1}{2}} \right] = - k \times 360
k=−ln⁡[2]−360k = \dfrac{{ - \ln \left[ 2 \right]}}{{ - 360}}
We know that ln⁡[2]=0.693\ln \left[ 2 \right] = 0.693
k=0.693360=0.0019min⁡k = \dfrac{{0.693}}{{360}} = 0.0019\min
When the radioactive substance has been decaying for 18hours18hours , then N0=36{N_0} = 36 grams
t=18hours=1080min⁡t = 18hours = 1080\min
k=0.0019min⁡k = 0.0019\min
So, ln⁡[NN0]=−kt\ln \left[ {\dfrac{N}{{{N_0}}}} \right] = - kt
⇒ln⁡[NN0]=−0.0019×1080\Rightarrow \ln \left[ {\dfrac{N}{{{N_0}}}} \right] = - 0.0019 \times 1080
⇒ln⁡[NN0]=−2.052\Rightarrow \ln \left[ {\dfrac{N}{{{N_0}}}} \right] = - 2.052
⇒NN0=ln⁡−1[−2.052]\Rightarrow \dfrac{N}{{{N_0}}} = {\ln ^{ - 1}}\left[ { - 2.052} \right]
We know that ln⁡−1[−2.052]=0.1284777{\ln ^{ - 1}}\left[ { - 2.052} \right] = 0.1284777
⇒N36=0.1284777\Rightarrow \dfrac{N}{{36}} = {\text{0}}{\text{.1284777}}
N=4.625gramN = 4.625gram
We have to calculate the amount of radioactive substance decayed, i.e. N0−N=36−4.625{N_0} - N = 36 - 4.625
N0−N=31.375gram{N_0} - N = 31.375gram
Therefore 31.375gram31.375gram of a radioactive substance decays in 1818 hours, if the half-life of that radioactive substance is 66 hours.

So, the correct answer is “Option A”.

Note:
To solve such types of problems in which we use the equation for the rate of radioactive decay, i.e. ln⁡[NN0]=−kt\ln \left[ {\dfrac{N}{{{N_0}}}} \right] = - kt , it is very important to know how to calculate the log values ( like in this question we used the value of ln⁡[2]\ln \left[ 2 \right] and ln⁡−1[−2.052]{\ln ^{ - 1}}\left[ { - 2.052} \right] ). Most of the time it is already given in the problem, but it was not in the given problem.