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Question: Two radioactive substances A and B have decay constants \(5\lambda \) and \(\lambda \) respectively....

Two radioactive substances A and B have decay constants 5λ5\lambda and λ\lambda respectively. At t=0t = 0 they have the same number of nuclei. The ratio of the number of nuclei of A to those of B will be (1e)2{\left( {\dfrac{1}{e}} \right)^2} after a time interval.
(A) 14λ\dfrac{1}{{4\lambda }}
(B) 4λ4\lambda
(C) 2λ2\lambda
(D) 12λ\dfrac{1}{{2\lambda }}

Explanation

Solution

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=N0eλtN = {N_0}{e^{ - \lambda t}}
Where, NN stands for the number of atoms at a given instant, N0{N_0}stands for the initial number of atoms, λ\lambda is called the decay constant or the disintegration constant and tt stands for the time

Complete step by step answer:
According to the law of radioactive disintegration, we can write the decay equation as
N=N0eλtN = {N_0}{e^{ - \lambda t}}
Let the number of atoms of A beNA{N_A}, its decay constant is given by 5λ5\lambda
Then we can write that the number of atoms of A is
NA=N0e5λt{N_A} = {N_0}{e^{ - 5\lambda t}}
Let the number of atoms of B beNB{N_B}, its decay constant is given by λ\lambda
Then we can write the number of atoms of B as
NB=N0eλt{N_B} = {N_0}{e^{ - \lambda t}}
Taking the ratio of NA{N_A}andNB{N_B}, we get
NANB=N0e5λtN0eλt=e4λt\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 NANB=1e2=e2\dfrac{{{N_A}}}{{{N_B}}} = \dfrac{1}{{{e^2}}} = {e^{ - 2}}
This means that, e4λt=e2{e^{ - 4\lambda t}} = {e^{ - 2}}
4λt=2\Rightarrow 4\lambda t = 2
t=24λ=12λ\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 (1e)2{\left( {\dfrac{1}{e}} \right)^2} after a time interval 12λ\dfrac{1}{{2\lambda }}

The answer is Option (D): 12λ\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. NN and N0{N_0} can be replaced by the mass of the material. The S.I. The unit of radioactivity is Becquerel (Bq).