A) first increases then decreasing
B) Goes on increasing
C) Goes on decreasing
D) Is \[\frac{K}{\lambda }\] after long time
Correct Answer: D
Solution :
| \[\xrightarrow[K\,nulei/sec]{\text{Production}}\,\,A\,\,\xrightarrow[\lambda ]{Decay}\,\,B\] |
| Rate of change of number nuclei of \[A=\frac{dN}{dt}\] |
| = Rate of formation of A- Rate of decay of \[A=K-\lambda N\] where N is number of nuclei of A at time t |
| \[\frac{dN}{dt}=K-\lambda N\Rightarrow \int\limits_{{{N}_{0}}}^{N}{\frac{dN}{K-\lambda N}}=\int\limits_{0}^{t}{dt}\] |
| \[N=\frac{K}{\lambda }(1-{{e}^{-\,{{\lambda }_{t}}}})+{{N}_{0}}{{e}^{-\,{{\lambda }_{t}}}}\] |
| After long time \[t=\infty \] number of nucleus will become \[\frac{K}{\lambda }\] |
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