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Consider radioactive decay of `A` to `B` with which further decays either to `X` or `Y`, `lambda_(1), lambda_(2)` and `lambda_(3)` are decay constant for `A` to `B` decay, `B` to `X` decay and Bto `Y` decay respectively. At `t=0`, the number of nuclei of `A,B,X` and `Y` are `N_(0), N_(0)` zero and zero respectively. `N_(1),N_(2),N_(3)` and `N_(4)` are the number of nuclei of `A,B,X` and `Y` at any instant `t`. At `t=oo`, which of following is incorrect ?A. `N_(2)=0`B. `N_(3)=(N_(0)lambda_(2))/(lambda_(2)+lambda_(3))`C. `N_(4)=(2N_(0)lambda_(3))/(lambda_(2)+lambda_(3))`D. `N_(1)+N_(2)+N_(3)+N_(4)=2N_(0)` |
Answer» Correct Answer - B At `t= oo` is `2N_(0)`. Also, the total number of nuclei at any instant remains the same `:. N_(1)+N_(2)+N_(3)+N_(4)=2N_(0)` The ratio of X and Y formed are in the ratio `lambda_(2): lambda_(3)` and the total number of nuclei of X and Y at `t=oo` is `2N_(0)`. |
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