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The half life of ""_(38) Sr^(90) isotope is 28 years. What is the rate of disintegration of 15 mg of this isotope? (Given Avogadro No =6.023 xx 10^(23) ) |
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Answer» Solution :`T_(1//2) = 28` years `= 28 XX 365 xx 24 xx 3600` `T_(1//2) = 8.83 xx 10^(8)` SECONDS `90g Sr_(38)^(90)` contains `6.0.23 xx 10^(23)` atoms `N = (6.023 xx 10^(23) xx 15 xx 10^(-3))/(90) = 1.004 xx 10^(20)` atoms `lambda = (0.693)/(T_(1//2)) = (0.693)/(8.83 xx 10^(8)) = 0.07848 xx 10^(-8)` `lambda = 7.85 xx 10^(-10)` Rateof disintegration `R = lambda N` `R = 7.85 xx 10^(-10) xx 1.004 xx 10^(20)` `R = 7.8814 xx 10^(10)B Q` |
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