Answer
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Hint: Radioactivity is the phenomenon of spontaneous disintegration of the atomic nucleus by the emission of highly penetrating radiations. The law of radioactive disintegration states that the rate of disintegration at any instant is directly proportional to the number of atoms of the element present at that instant.
Formula used
$N = {N_0}{e^{ - \lambda t}}$
Where, $N$ stands for the number of atoms at a given instant, ${N_0}$stands for the initial number of atoms, $\lambda $is called the decay constant or the disintegration constant and $t$ stands for the time
Complete step by step answer:
According to the law of radioactive disintegration, we can write the decay equation as
$N = {N_0}{e^{ - \lambda t}}$
Let the number of atoms of A be${N_A}$, its decay constant is given by $5\lambda $
Then we can write that the number of atoms of A is
${N_A} = {N_0}{e^{ - 5\lambda t}}$
Let the number of atoms of B be${N_B}$, its decay constant is given by $\lambda $
Then we can write the number of atoms of B as
${N_B} = {N_0}{e^{ - \lambda t}}$
Taking the ratio of ${N_A}$and${N_B}$, we get
$\dfrac{{{N_A}}}{{{N_B}}} = \dfrac{{{N_0}{e^{ - 5\lambda t}}}}{{{N_0}{e^{ - \lambda t}}}} = {e^{ - 4\lambda t}}$
In the question, it is given that $\dfrac{{{N_A}}}{{{N_B}}} = \dfrac{1}{{{e^2}}} = {e^{ - 2}}$
This means that, ${e^{ - 4\lambda t}} = {e^{ - 2}}$
$\Rightarrow 4\lambda t = 2$
$\Rightarrow t = \dfrac{2}{{4\lambda }} = \dfrac{1}{{2\lambda }} $
So, the ratio of number of nuclei of A to those of B will be ${\left( {\dfrac{1}{e}} \right)^2}$ after a time interval $\dfrac{1}{{2\lambda }}$
The answer is Option (D): $\dfrac{1}{{2\lambda }}$
Note
The disintegration constant represents the probability of an atom to disintegrate. The negative sign in the disintegration constant indicates that the number of atoms decreases with the increase in time. The number of un-disintegrated atoms of a radioactive substance decreases exponentially. $N$ and ${N_0}$ can be replaced by the mass of the material. The S.I. The unit of radioactivity is Becquerel (Bq).
Formula used
$N = {N_0}{e^{ - \lambda t}}$
Where, $N$ stands for the number of atoms at a given instant, ${N_0}$stands for the initial number of atoms, $\lambda $is called the decay constant or the disintegration constant and $t$ stands for the time
Complete step by step answer:
According to the law of radioactive disintegration, we can write the decay equation as
$N = {N_0}{e^{ - \lambda t}}$
Let the number of atoms of A be${N_A}$, its decay constant is given by $5\lambda $
Then we can write that the number of atoms of A is
${N_A} = {N_0}{e^{ - 5\lambda t}}$
Let the number of atoms of B be${N_B}$, its decay constant is given by $\lambda $
Then we can write the number of atoms of B as
${N_B} = {N_0}{e^{ - \lambda t}}$
Taking the ratio of ${N_A}$and${N_B}$, we get
$\dfrac{{{N_A}}}{{{N_B}}} = \dfrac{{{N_0}{e^{ - 5\lambda t}}}}{{{N_0}{e^{ - \lambda t}}}} = {e^{ - 4\lambda t}}$
In the question, it is given that $\dfrac{{{N_A}}}{{{N_B}}} = \dfrac{1}{{{e^2}}} = {e^{ - 2}}$
This means that, ${e^{ - 4\lambda t}} = {e^{ - 2}}$
$\Rightarrow 4\lambda t = 2$
$\Rightarrow t = \dfrac{2}{{4\lambda }} = \dfrac{1}{{2\lambda }} $
So, the ratio of number of nuclei of A to those of B will be ${\left( {\dfrac{1}{e}} \right)^2}$ after a time interval $\dfrac{1}{{2\lambda }}$
The answer is Option (D): $\dfrac{1}{{2\lambda }}$
Note
The disintegration constant represents the probability of an atom to disintegrate. The negative sign in the disintegration constant indicates that the number of atoms decreases with the increase in time. The number of un-disintegrated atoms of a radioactive substance decreases exponentially. $N$ and ${N_0}$ can be replaced by the mass of the material. The S.I. The unit of radioactivity is Becquerel (Bq).
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