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An experiment requires minimum beta-activity produced at the rate of 346 beta-particles per minute. The t_((1)/(2)) " of" ""_(42)^(99)No, which is a beta-emitter, is 66.6 hours. Find the minimum amount of ""_(42)^(99)Mo required to carry out the experiment in 6.909 hours. |
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Answer» Solution :`beta` activity `= -(d(N))/(dt)= 346 xx 60 beta`-particles/h `therefore -(d(N))/(dt)= lamda (N)= (0.6932)/(t_((1)/(2))) xx N` or `346 xx 60= (0.6932)/(66.6) N` or `N= 2 xx 10^(6)` Again, `lamda= (2.303)/(t) "LOG" (N^(0))/(N)` `therefore (0.6932 )/(66.6) = (2.303)/(6.909) "log" (N^(0))/(N)= (1)/(3) "log" (N^(0))/(N)` SUBSTITUTING the VALUE of N in the above equation, log `(N^(0))/(N)= (3 xx 0.6932)/(66.6)= 0.0312` or `(N^(0))/(N)= 1.074` or `N^(0)= 1.074 xx N = 1.074 xx 2 xx 10^(6)= 2.148 xx 10^(6)` `therefore` no. of moles of `Mo= (2.148 xx 10^(6))/(6.022 xx 10^(23))= 3.567 xx 10^(-18)` and wt. of `Mo= (3.567 xx 10^(-18) xx 99)g` (Mo= 99) `=3.53 xx 10^(-16)g` |
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