Answer:
0.835 mg of a 2.000-mg sample remains after 6.55 years.
Step-by-step explanation:
We are given that The half-life of cobalt-60 is 5.20 yr.
Formula of half life : [tex]N(t) = N(0)e^{\frac{-0.693t}{t\frac{1}{2}}}[/tex]
Where N(0) = Initial amount
N(t) = Quantity remaining after time
t = time
[tex]\frac{t}{2} =\text{half life}[/tex]
So, N(0) = Initial amount = 2 mg
t = time = 6.55 years
[tex]\frac{t}{2} =\text{half life} = 5.20 yr [/tex]
Substitute the values in the formula
Therefore,
[tex]N(t) = 2 e^{\frac{-0.693 \times 6.55}{5.20}}[/tex]
[tex]N(t) =0.835[/tex]
Hence 0.835 mg of a 2.000-mg sample remains after 6.55 years.
