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The rate of radioactive decay of a sample are `3 xx 10^(8) dps` and `3 xx 10^(7) dps` after time `20 min` and `43.03 min` respectively. The fraction of radio atom decaying per second is equal toA. `A. (1)/(600)`B. `B. 1`C. `C. 0.5`D. `D. 0.001` |
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Answer» Correct Answer - A `k=(2.303)/(t_(2)-t_(1))log. (r_(1))/(r_(2))=(2.303)/(23.03)log.(3xx10^(8))/(3xx10^(7))= 0.1min^(-1)` `k = -(dN//N)/(dt) = 0.1 min^(-1) = (1)/(600)s^(-1)` |
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