The radioactive isotope `.^(137)Cs` has a half life period of 30 years. Staring with 1 mg of `.^(137)Cs`. How much would remain after 120 years?

Explanation

At this time, we have 1.0 mg of `.^(137)Cs`, after 30 years, we shall have one half of the original, or 0.50 mg after 60 years, we shall have 0.25 mg after, after 90 years, we shall have 0.125 mg and finally, after 120 years we shall have 0.625 mg
after `{:(30 "years",-,0.050 mg),(60 "years",-,0.25 mg),(90 "years",-,0.125 mg),(120 "years",-,0.0625 mg):}`
Alternative solution : Total time = 120 years
total time `= n xx t_(1//2)`
`120 = n xx 30`
n = 4
Thus, the quantity of the isotope left after
four half life periods `= ((1)/(2))^(4) N_(0) = ((1)/(2))^(4) xx 1`
`= (1)/(16) = 0.0625 mg`