Question

Nuclei $ A $ and $ B $ convert into a stable nucleus $ C $ . Nucleus $ A $ is converted into $ C $ by emitting $ 2\alpha - $ particles and $ 3\beta - $ particles. Nucleus $ B $ is converted into $ C $ by emitting one $ \alpha - $ particle and $ 5\beta - $ particles. At time $ t = 0 $ , nuclei of $ A $ are $ 4{N_ \circ } $ ​ and nuclei of $ B $ are $ {N_ \circ } $ ​. Initially, the number of nuclei of $ C $ are zero. Half-life of $ A $ (into conversion of $ C $ ) is $ 1\min $ and that of $ B $ is $ 2\min $ . Find the time $ \left( {{\text{in}}\,\,{\text{minutes}}} \right) $ at which rate of disintegration of $ A $ and $ B $ are equal.

Answer 1

Hint: Here in this question for solving it we will use the order decay and we know that the first order decay is given by $ \dfrac{{dA}}{{dt}} = - {\lambda _A}A $ and for $ B $ , the first order decay will be $ \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,
 $ A \to C $ and $ B \to C $
Now the first order decay will be given as,
 $ \Rightarrow \dfrac{{dA}}{{dt}} = - {\lambda _A}A $
And on solving for the value of $ A $ , we get
 $ \Rightarrow A = 4{N_ \circ }{e^{ - {\lambda _A}A}} $
And for $ B $ , first order decay will be $ \dfrac{{dB}}{{dt}} = - {\lambda _B}B $
And on solving for the value of $ B $ , we get
 $ \Rightarrow B = 4{N_ \circ }{e^{ - {\lambda _B}B}} $
Therefore the rate of disintegration will become
 $ \Rightarrow \dfrac{{dA}}{{dT}} = \dfrac{{dB}}{{dT}} $
Now on substituting the values, we get
 $ \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
 $ \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
 $ \Rightarrow t = 6\min $
Therefore, the time $ \left( {{\text{in}}\,\,{\text{minutes}}} \right) $ at which rate of disintegration of $ A $ and $ B $ are equal to $ 6\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.