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Element X is a radio isotope with half lives of 11 yrs. The initial mass of element x is 60 grams. How much to the nearest tenth of the element remains after 25 yrs?

User RavensKrag
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We are given that a radio isotope has an initial mass of 60 grams and a half-life of 11 years. The mass of the isotope with respect to time can be modeled using the following exponential function:


M=m_0e^(kt)

Where:


\begin{gathered} M=\text{ mass} \\ m_0=\text{ initial mass} \\ k=\text{ constant} \\ t=\text{ time} \end{gathered}

Now, since we are given that the half-life is 11 years this means that the mass will be half of the initial mass when "t = 11 years". We can use this fact to calculate the value of "k". Substituting half the initial mass we get:


(m_0)/(2)=m_0e^{k(11\text{year)}}

Now, we may cancel the initial mass:


(1)/(2)=e^{k(11\text{years)}}

Now, we solve for "k". To do that we will take natural logarithm to both sides:


\ln (1)/(2)=\ln e^{k(11\text{years)}}

Now, we use the following property of logarithms:


\ln x^y=y\ln x

Applying the property we get:


\ln (1)/(2)=k(11years)\ln e^{}

We have that:


\ln e=1

Therefore:


\ln (1)/(2)=k(11years)

Now, we divide both sides by 11:


(1)/(11)\ln (1)/(2)=k

Solving the operations:


-0.063=k

Now, we substitute in the formula for the mass:


M=60e^(-0.063t)

Now, to determine the value of the mass after 25 years we substitute the value "t = 25", we get:


M=60e^(-0.063(25))

Solving the operations:


M=12.4

Therefore, the mass after 25 years is 12.4 grams.

User Marty Wallace
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