During nuclear explosion one of the products is ""^(90)Sr with half-life of 28.1 years .If 1 mug of ""^(90)Sr was absorbed in the boned of a newly born baby instead of calcium ,how much of it wil reman after 10 years and 60 years if it is not lost metabolically.

During nuclear explosion one of the products is ""^(90)Sr with half-li

1.	During nuclear explosion one of the products is ""^(90)Sr with half-life of 28.1 years .If 1 mug of ""^(90)Sr was absorbed in the boned of a newly born baby instead of calcium ,how much of it wil reman after 10 years and 60 years if it is not lost metabolically.
Answer» Solution :Nuclear explosion is a first order reaction. `1 mug ""^(90)Sr` is observed in NEW born baby. `therefore[R]_(0)=1mug` Time `t_(1)` =fter 10 years,so,`[R]_(t_(2))`=y `mug` Constant of explosion reaction =k.If the reaction is of first order. `k=(0.693)/(t_((1)/(2)))=(0.693)/(28.1 year)=2.4662xx10^(-2)years` After 10 years ,the calculation of remianing `""^(90)Sr`: `t=(2.303)/(k)` log `([R]_(0))/([R]_(t))` where ,t=10 `therefore 10=(2.303)/(2.4662xx10^(-2))(log 1 mug)/(x mug)` `therefore (1)/(x)=(10xx0.024662)/(2.303)=0.1070 ` `therefore (1)/(x)`=ANTI log 0.1070 `therefore (1)/(2)=1.2794` `therefore=0.7816 mu g ""^(90)Sr` will be remaining. Suppose `""^(90)S=y mug` after t=60 year, `t=(2.303)/(k)` log `([R]_(0))/([R]_(r))` `therefore 60=(2.303)/(0.024662)` log`(1)/(y)` `therefore (60xx0.024662)/(2.303)` =log `-log y `therefore` 0.6425=0-log y=-log y `therefore` log y=-0.6425 `therefore` y=Antilog (-0.6425)