Determine how much will remain algebra help needed

Question

The half-life of Polonium-209 is 102 years.

If we start with a sample of 108 mg of Polonium-209, determine how much will remain after 153 years.

If necessary, round answer to three decimal places.

___mg

Answer 1

Start with the initial amount, 108 mg. After one "half life" there will be 54 mg, right? After two "half-lifes" there will be 27 mg, right?

So the amount you have, A(t), equals the initial amount multiplied by (1/2) raised to the "number of half-lifes" power, if that makes sense. To find the "number of half-lifes" take the number of years that go by divided by 102. So if 102 years go by, it will be 102/102 or 1 half-life. If 204 years go by, it will be 204/102 or 2 "half-lifes".

So:

A(t) = 108(1/2)^t/102

Meaning that the amount after 153 years is 108(1/2)^153/102 = 108(1/2)^1.5 = 108(0.35355339)

I'm sure you can do the rest in your calculator.

Answer 2

The half-life formula is N(t) = N₀(1/2)^t/t_(1/2) where:

N(t) = quantity of the substance remaining <-- what we need to determine

N₀ = initial quantity of the substance = 108 mg

t = time elapsed = 153 years

t_1/2 = half-life of the substance = 102 years

N(t) = N₀(1/2)^t/t_(1/2) = 108(1/2)^153/102 ≈ 38.184

Therefore, there will be 38.184 mg of the sample left.

Hope this helped!