A radioisotope has half life of 10 years. What percentage of the original amount of it would you expect to remain after 20 years? a)0 b)12.5 c)25 d)8

Options

A. 0
B. `12*5`
C. 25
D. 8

Answer

C

Explanation

`C_(n)=(C^(@))/(2^(n))`
`C_(n)` =Conc. After n half lives.
`C^(@)`=initial conc.
`C_(n)` =Conc. After n half lives
`C^(@)`= initial conc.
n= no. of half lives
20 years =2 half lives
`therefore C_(n)=(1)/(2^(2))=(1)/(4)=0.25` or 25%