InterviewSolution
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InterviewSolution
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Two radioactive materials `X_1` and `X_2` have decay constants `10 lamda` and `lamda` respectively. If initially they have the same number of nuclei, then the ratio of the number of nuclei of `X_1` to that of `X_2` will be `1//e` after a time.A. `lamda`B. `(1)/(2) lamda`C. `(1)/(4 lamda)`D. `( e)/(lamda)` |
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Answer» Correct Answer - C ( c) If `N` is the number of radioactive nuclei present at some instant, then `N = N_0 e^(-lamda t)` The constant `N_0` represents the number of radioactive nuclei at `t = 0` Now, `(N_1)/(N_2) = (e^(-lamda_1 t))/(e^(-lamda_2 t))`or `(N_1)/(N_2) = (e^(-5 lamda t))/(e^(-lamda t)) = e^(-4 lamda t)` but `(N_1)/(N_2) = (1)/( e)` (as provided) Therefore, `(1)/( e) = (1)/(e^(4 lamda t))` or `4 lamda t = 1` or ` t = (1)/(4 lamda)`. |
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