Question

The half-life of cobalt-60 is 5.20 yr. how many milligrams of a 2.000-mg sample remains after 6.55 years?

Answer 1

Decay expression is stated as follows:

N(t) = N(o)e^(-0.693t/t1/2)
N(t) = Quantity remaining after time, t; N(o) = Initial quantity, t = time, t1/2 = Half life

Therefore,
N(6.55) = 2*e^(-0.693*6.55/5.2) = 0.835 g

Answer 2

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 : $N(t) = N(0)e^{\frac{-0.693t}{t\frac{1}{2}}}$

Where N(0) = Initial amount

N(t) = Quantity remaining after time

t = time

$\frac{t}{2} =\text{half life}$

So, N(0) = Initial amount = 2 mg

t = time = 6.55 years

$\frac{t}{2} =\text{half life} = 5.20 yr$

Substitute the values in the formula

Therefore,

$N(t) = 2 e^{\frac{-0.693 \times 6.55}{5.20}}$

$N(t) =0.835$

Hence 0.835 mg of a 2.000-mg sample remains after 6.55 years.