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.

Answer and Explanation:

{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}


Learn more about this topic:

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Calculating Rate and Exponential Growth: The Population Dynamics Problem

from Math 104: Calculus

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