A sample of 500 g of radioactive lead-210 decays to polonium-210 according to the function A(t)=500e^-0.032t , where t is time in years. Find the amount of radioactive lead remaining after (a) 3yr, (b) 8yr, (c) 10 yr. (d) Find the half-life.

Step-by-step explanation:

Using the equation A(t) = 400e-.032t

a) replace t with 4 so A(4) = 400e((-.032)(4))

The hardest part about this is making sure to use order of operations. Be certain it works like this:

A(4) = 400e-.128

A(4) = 400(.8799)

A(4) = 351.9 grams

b) A(8) = 400e((-.032)(8)) = 309.7 grams

c) A(20) = 400e((-.032)(20)) = 210.9 grams

Note here that even after 20 years, not quite half of the original amount is gone. So, we can anticipate that in finding the half life, that our answer should be slightly greater than 20 years.

d) 200 = 400e(-.032t)

Divide both sides of the equation by 400.

.5 = e(-.032t)

Change this to logarithmic form.

Ln .5 = -.032t

-.6931≈ -.032t

t ≈ 21.7 years

Hope this helps!

the reporter's right

step-by-step explanation:

1/4 = .25 / 25%

a

step-by-step explanation:

the answer is a