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Question: Nuclei \( A \) and \( B \) convert into a stable nucleus \( C \) . Nucleus \( A \) is converted into...

Nuclei AA and BB convert into a stable nucleus CC . Nucleus AA is converted into CC by emitting 2α2\alpha - particles and 3β3\beta - particles. Nucleus BB is converted into CC by emitting one α\alpha - particle and 5β5\beta - particles. At time t=0t = 0 , nuclei of AA are 4N4{N_ \circ } ​ and nuclei of BB are N{N_ \circ } ​. Initially, the number of nuclei of CC are zero. Half-life of AA (into conversion of CC ) is 1min1\min and that of BB is 2min2\min . Find the time (inminutes)\left( {{\text{in}}\,\,{\text{minutes}}} \right) at which rate of disintegration of AA and BB are equal.

Explanation

Solution

Here in this question for solving it we will use the order decay and we know that the first order decay is given by dAdt=λAA\dfrac{{dA}}{{dt}} = - {\lambda _A}A and for BB , the first order decay will be dBdt=λBB\dfrac{{dB}}{{dt}} = - {\lambda _B}B . And from this rate of integration will be calculated and solved for the value of the time, we will get the answer.

Complete step by step answer
As here in this question the conversion takes place like,
ACA \to C and BCB \to C
Now the first order decay will be given as,
dAdt=λAA\Rightarrow \dfrac{{dA}}{{dt}} = - {\lambda _A}A
And on solving for the value of AA , we get
A=4NeλAA\Rightarrow A = 4{N_ \circ }{e^{ - {\lambda _A}A}}
And for BB , first order decay will be dBdt=λBB\dfrac{{dB}}{{dt}} = - {\lambda _B}B
And on solving for the value of BB , we get
B=4NeλBB\Rightarrow B = 4{N_ \circ }{e^{ - {\lambda _B}B}}
Therefore the rate of disintegration will become
dAdT=dBdT\Rightarrow \dfrac{{dA}}{{dT}} = \dfrac{{dB}}{{dT}}
Now on substituting the values, we get
4λAeλAt=λBeλBt\Rightarrow 4{\lambda _A}{e^{ - {\lambda _A}t}} = {\lambda _B}{e^{ - {\lambda _B}t}}
And on solving the above equation, we get the equation as
4ln21eln21t=ln22eln22t\Rightarrow 4\dfrac{{\ln 2}}{1}{e^{ - \dfrac{{\ln 2}}{1}t}} = \dfrac{{\ln 2}}{2}{e^{ - \dfrac{{\ln 2}}{2}t}}
And on solving for the time, we will get
t=6min\Rightarrow t = 6\min
Therefore, the time (inminutes)\left( {{\text{in}}\,\,{\text{minutes}}} \right) at which rate of disintegration of AA and BB are equal to 6min6\min .

Note:
In radioactivity, the rate of decay turns out to be proportional to the present number of particles at any time. Decay rate is equal to the number of particles and the product with the decay constant. The decay constant is the probability of a nucleus decaying in unit time. Hence, the decay rate is directly proportional to the decay constant.