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Determine the mass of Na^(22) which has an activity of 5mCi. Half life of NA^(22) is 2.6 years. Avogadro number =6.023xx10^(23) atoms. |
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Answer» SOLUTION :Activity of `Na^(22) , A= (dN)/( dt) =5 m Ci` `A= 5 XX 10^(-3) xx 3.7 xx 10^(10)` `A=1.85 xx 10^(+7)` disintegration per second Disintegration constant `lambda = (0.693)/(T)= (0.693)/( 2.6 xx 365 xx 24 xx 60 xx 60)` `lambda = 8.453 xx 10^(-9) s^(-1)` Activity, `(d N)/( dt) = lambda N` `N= (1)/( lambda) (dN)/( dt)` `N= 2.189 x 10^(16)` atoms According to Avagadro.s hypothesis `6.023 xx 10^(23)` atoms of `Na^(22)` weigh 22 gm `2.189 xx 10^(16)` atoms weigh `= (22 xx 10^(-3))/( 6.023 xx 10^(23)) xx 2.189 xx 10^(16)` `=7.996 xx 10^(-10)` kg |
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