InterviewSolution
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InterviewSolution
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A solution of 1 litre has `0.6g` of non-radioactive `Fe^(3+)` with mass no. 56. To this solution `0.209g` of radioactove `Fe^(2+)` is added with mass no. 57 and the following reaction occurred. `.^(57)Fe^(2+) + .^(56)Fe^(3+) rarr .^(57)Fe^(3+) + .^(56)Fe^(2+)` At the end of one hour it was found that `10^(-5)` moles of non-radioactive `.^(56)Fe^(2+)` mol `L^(-1) hr^(-1)`. Negalecting any charge in volume, calculate the activity of the sample at the end of `1 hr` (`t_(1//2)` for `.^(57)Fe^(2+) = 4.62 hr`.) |
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Answer» `{:(,.^(57)Fe^(2+)+,.^(56)Fe^(3+)to,.^(57)Fe^(3+)+,.^(56)Fe^(2+)),("Before reaction",0.209/57,0.6/56,0,0),("After reaction",[0.209/57-10^(-5)],[0.6/56-10^(-5)],10^(-5),10^(-5)):}` `.^(57)Fe^(2+)` left after reaction `= 3.667xx10^(-3) - 10^(-5)` `= 366.6xx10^(5)` mole or `N_(0) = 366.6xx10^(-5)xx6.023xx10^(23) = 2.208xx10^(21)` Also `t = (2.303)/(lambda) log (N_(0))/(N)` `1 = (2.303xx4.62)/(0.693) log (2.208xx10^(21))/(N)` `N = 1.9xx10^(21)` `:.` Rate of decay `= lambda xx N = (0.693)/(4.62) xx 1.9xx10^(21)` `= 2.85xx10^(20)` dph |
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