Category : JEE Main & Advanced
Suppose a radioactive element A disintegrates to form another radioactive element B which intern disintegrates to still another element C; such decays are called successive disintegration.
Rate of disintegration of \[A=\frac{d{{N}_{1}}}{dt}=-{{\lambda }_{1}}{{N}_{1}}\] (which is also the rate of formation of B)
Rate of disintegration of \[B=\frac{d{{N}_{2}}}{dt}=-{{\lambda }_{2}}{{N}_{2}}\]
\[\therefore \] Net rate of formation of B = Rate of disintegration of A - Rate of disintegration of B
\[={{\lambda }_{1}}{{N}_{1}}-{{\lambda }_{2}}{{N}_{2}}\]
Equilibrium
In radioactive equilibrium, the rate of decay of any radioactive product is just equal to it's rate of production from the previous member.
i.e.\[{{\lambda }_{1}}{{N}_{1}}={{\lambda }_{2}}{{N}_{2}}\Rightarrow \]\[\frac{{{\lambda }_{1}}}{{{\lambda }_{2}}}=\frac{{{N}_{2}}}{{{N}_{2}}}=\frac{{{\tau }_{2}}}{{{\tau }_{1}}}=\frac{({{T}_{1/2}})}{{{({{T}_{1/2}})}_{1}}}\]
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