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# 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 $$A$$ and $$4{N}_{0}$$ and nuclei of $$B$$ are $${N}_{0}$$. Initially, number of nuclei of $$C$$ are zoer. Half-life of $$A$$ (into conversion of $$C$$) is $$1min$$ and that of $$B$$ is $$2min$$. Find the time (in minutes) at which rate of disintegration of $$A$$ and $$B$$ are equal.

A
$$6$$
Solution
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#### Correct option is A. $$6$$We have to find the time at which$${ \lambda }_{ A }{ Nu }_{ A }={ \lambda }_{ B }{ Nu }_{ B }$$$$\left( \cfrac { \ln { 2 } }{ { T }_{ A } } \right) \left( 4{ N }_{ 0 }{ e }^{ -{ \lambda }_{ A }t } \right) =\left( \cfrac { \ln { 2 } }{ { T }_{ B } } \right) \left(N_0 { e }^{ -{ \lambda }_{ B }t } \right) \quad \quad$$$$\Rightarrow{ e }^{ ({ \lambda }_{ A }-{ \lambda }_{ B })t }=8$$$$\Rightarrow ({ \lambda }_{ A }-{ \lambda }_{ B })t=\ln { 8 } =3\left( \ln { 2 } \right)$$$$\left( \cfrac { \ln { 2 } }{ 1 } -\cfrac { \ln { 2 } }{ 2 } \right) t=3\ln { \left( 2 \right) }$$$$\Rightarrow t=6min$$

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