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(a) Write symbolically the beta^(-) decay process of " "_(15)^(32)P. (b) Derive an expression for the average life of a radionuclide. Give its relationship with the half-life. |
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Answer» Solution :(a) The `BETA^(-)` decay process of `" "_(15)^(32)P` is symbolically written as : `" "_(15)^(32)Pto" "_(16)^(32)S+" "_(-1)^(0)e+bar(nu)` Here `e^(-)` is the `beta^(-)` PARTICLE and `bar(nu)` is antineutrino particle. (b) Let in a given radioactive sample INITIALLY we have `N_(0)` nuclei. At time t the number of nuclei intact is `N= N_(0)e^(-lambdat)`. The number of nuclei which decay in the time interval f to t + dt is `(-(dN)/(dt)).dt=lambdaNdt=lambdaN_(0)e^(-lambdat)dt`. Each of them has lived for time t. Hence, the total life of all these nuclei would be `tlambdaN_(0)e^(-lambdat)dt`. To obtain the average life of a RADIONUCLIDE we have to integrate this expression over all times from 0 to `oo` and divide by the total number `N_(0)` of nuclei at the beginning. `therefore` Average life period `tau = (lambdaN_(0)int_(0)^(oo)(t.e^(-lambdat)dt))/N_(0)=lambdaint_(0)^(oo)(te^(-lambdat)dt)=lambda[{(t.e^(-lambdat))/(-lambda)}_(0)^(oo)-int_(0)^(oo)(e^(-lambdat)/(-lambda)dt)]=lambda[(t.e^(-lambdat))/(-lambda)-e^(-lambdat)/(lambda)^(2)]_(0)^(oo)=1/(lambda)` `therefore` Half-life period `T_(1/2) = (ln2)/(lambda)= 0.693/(lambda)=0.693tau` |
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