Question

During nuclear explosion, one of the products is `.^(99)Sr` with half`-` life of `28.1 `years. If `1 mug` of `.^(90)Sr` was absorbed in the bones of a newly born baby instead of calcium, how much of its will remain after 10 years and 60 years if it is not lost metabolically.

Answer 1

`k = (0.693)/(t_((1)/(2))) = (0.693)/(28.1) y^(-1)`
Here ,
It is known that ,
`t = (2.303)/(k) "log" ([R]_(0))/([R])`
`implies 10 = (2.303)/((0.693)/(28.1)) "log" (1)/([R])`
`implies 10 = (2.303)/((0.693)/( 28.1)) (-"log"[R])`
`implies "log"[R] = - (10 xx 0.693)/(2.303 xx 28.1)`
`implies [R]` = antilog `(-0.1071`)
= antilog `(bar1 . 6=8929)`
= `0.7814 mu g `
Therefore , `0.7814 mu g ` of `"^(90)Sr` will remain after 10 years .
Again ,
`t = (2.303)/(k) "log" ([R]_(0))/([R])`
`implies 60 = (2.303)/((0.693)/(28.1)) "log" (1)/([R])`
`implies "log" [R] = -(60 xx 0.693)/(2.303 xx 28.1)`
`implies [R]` = antilog `(-0.6425)`
= antilog (`bar1 . 3575)`
= `0.2278 mu g`
Therefore , `0.2278 mu g` of `"^(90)Sr` will remain after 60 years .