Question

The radioactive element polonium decays according to the law given below where Q0 is the initial amount and the time t is measured in days. Q(t) = Q0 · 2-(t/140) If the amount of polonium left after 560 days is 80 mg, what was the initial amount present?

Answer 1

Answer:

1280 mg

Explanation:

Radioactive  decay is a phenomenon that occurs when a certain isotope of an element, said to be radioactive, decays, turning into a lighter nucleus and emitting radiation + energy in the process.

The radioactive decay of this isotope of polonium is described by the equation

$Q(t)=Q_0 2^{-\frac{t}{140}}$

where

$Q(t)$ is the amount of polonium left after time t

$Q_0$ is the amount of polonium t time t = 0

$140$ is the half-life of the polonium, in days (it is the time it takes for the initial element to halve its amount)

IN this problem, we know that:

After t = 560 days, the amount of polonium left is $Q(t)=80 mg$. Therefore, we can re-arrange the equation, substituting t = 560 d, and solve for $Q_0$ to find the initial amount of polonium:

$Q_0 = Q(t) \cdot 2^{\frac{t}{140}}=(80)\cdot 2^{\frac{560}{140}}=1280 mg$