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If alpha=2N_0lambda, calculate the number of nuclei of A after one half-life of A, and also the limiting value of N as |
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Answer» `2N_0, 5/2N_0` `therefore N=1/lambda[2lambdaN_0-(2lambdaN_0-lambdaN_0)e^(-lambdat)]` or `N=(lambdaN_0)/lambda[2-e^(In(2))][Here""e^(-In(2))=2^(-1) = 1/2]` or `N=(lambdaN_0)/lambda[2-1/2] =(3N_0)/2 ` or `N=3/2N_0` When `t to oo ` and `alpha = 2lambdaN_0` `N=alpha/lambda=(2lambdaN_0)/lambda=2N_0` or `N=2N_0` |
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