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A radioactive substance has 6.0 xx 10^18 active nuclei initially. What time is required for the active nuclei o the same substance to become 1.0 xx 10^18 if its half-life is 40 s. |
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Answer» SOLUTION :The NUMBER of active NUCLEI at any instant of time t, `N_0/N=e^(lambdat)` `log_e (N_0/N)=lambdat` `therefore t=(log_e (N_0/N))/lambda=(2.303 log_10 (N_0/N))/lambda` In this PROBLEM , the initial number of active nuclei, `N_0=6.0xx10^18` `N=1.0xx10^18`, T=40 s `lambda=0.693/T=0.693/40=1.733xx10^(-2) s^(-1)` `t=(2.303 log_10 ((6.0xx10^18)/(1.0xx10^18)))/(1.733xx10^(-2)` `=(2.303 log_10 (6))/(1.733xx10^(-2))=(2.303xx0.7782)/(1.733xx10^(-2))`=103.4s |
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