# The radioactive element polonium decays according to the law given below where Q0 is the initial...

## Question:

The radioactive element polonium decays according to the law given below where {eq}Q_0 {/eq} is the initial amount and the time t is measured in days. {eq}Q(t)={{Q}_{0}}\times {{2}^{-(\frac{t}{140})}} {/eq} If the amount of polonium left after 700 days is 45 mg, what was the initial amount present?

## Exponential Rate:

{eq}\\ {/eq}

Mathematical analysis of the radioactive decay of substance is generally observed as an exponential process and hence involves powers of sum fixed numbers in the equation of decay.

{eq}\\ {/eq}

Given : The initial amount of radioactive element present {eq}=Q_{0} \ mg. {/eq}

Time of decay {eq}=700 {/eq} days.

Quantity of Polonium left after {eq}700 {/eq} days {eq}=Q_{700}=45 \ mg. {/eq}

Using the formula {eq}Q(t)={{Q}_{0}}\times {{2}^{-(\frac{t}{140})}} {/eq}, we get :-

{eq}\Rightarrow 45=Q_{0}\times 2^ {\left(-\frac{700}{140}\right)}\\\Rightarrow Q_{0}=45\times 2^5\\\Rightarrow \boxed{Q_{0}=1440 \ mg.} {/eq} 