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Two radioactive substances X and Y initially contain equal number of atoms . Their half-lives are 1 hour and 2 hours respectively . Calculate the ratio of their rates of disintegrations after four hours. |
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Answer» Solution :Rate of disintegration `(R),(DN)/(dt)= lamdaN = N XX (0.693)/T` `:.` The ratio of rates disintegration , `R_1/R_2=(N_1/N_2)T_2/T_1` Here , `T_(1) = 1 ` hour : `T_2` = 2 hours `N_0` is same for both radio active substances. In four hours , X completes 4 half - lives as its half - life is 1 hour. The REMAINING number of ATOMS of X, `N_(1)= N_0/(2^n)= N_0/(2^4) = N_0/16` . Y completes 2 half - lives in four hours as its half - life 2 hours. The number of atoms of Y remaining . `N_2=N_0/2^2=N_0/4"" :. R_1/R_2=N_0/16xx4/N_0(2/1)=1/2` |
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