Answer:
800 milligrams strontium-90 sample was present in the beginning.
Explanation:
Formula used :
[tex][A]=\frac{[A_o]}{2^n}[/tex]
where,
Amount of sample left after n-half lives = [A]
Initial amount of the sample = [tex][A_o][/tex]
n = number of half lives
We have:
Final amount of strontium-90 left after 5 half lives=[A]=25 mg,
Initial amount of the strontium-90 sample = [tex][A_o]=?[/tex]
n = 5
[tex][A]=\frac{[A_o]}{2^n}[/tex]
[tex]25 mg=\frac{[A_o]}{2^{5}}[/tex]
[tex][A_o]=25mg\times 2^5=800 mg[/tex]
800 milligrams strontium-90 sample was present in the beginning.
