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At time $t = 0, N_1$ nuclei of decay constant $\lambda _1 \,\& \,N_2$ nuclei of decay constant $\lambda _2$ are mixed . The decay rate of the mixture is :
${N_1}{N_2}{e^{ - \left( {{\lambda _1} + {\lambda _2}} \right)t}}$
$ + \left( {\frac{{{N_1}}}{{{N_2}}}} \right){e^{ - \left( {{\lambda _1} - {\lambda _2}} \right)t}}$
$ + ({N_1}{\lambda _1}{e^{ - {\lambda _1}t}} + {N_2}{\lambda _2}{e^{ - {\lambda _2}t}})$
$ + {N_1}{\lambda _1}{N_2}{\lambda _2}{e^{ - \left( {{\lambda _1} + {\lambda _2}} \right)t}}$
Solution
Decay rate of one particle only depends on its number od atoms but not on the other particles.So the total rate will be the addition of the rates of two compounds.
No. of atoms of first compound at any time $-N_{1} e^{-d_{1} t}$
No. of atoms of $2^{\text {nd }}$ compound at any time $=N_{1} e^{-\lambda_{2} t}$
we know that $\frac{d N}{d t}=\lambda N$
decay rate of $1^{st}$ compound $=\lambda_{1} N_{1} e^{-\lambda_{1} t}$
decay rate of $2^{\text {nd }}$ compound $=\lambda_{2} N_{2} e^{-\lambda_{2} t}$
Total rate of mixture $=\lambda_{1} N_{1} e^{-\lambda_{1} t}+\lambda_{2} N_{2} e^{-\lambda_{2} t}$
