Element $\mathrm{X}$ is a radioactive isotope such that every $11$ years, its mass decreases by half. Given that the initial mass of a sample of Element $X$ is $5600$ grams, how much of the element would remain after $10$ years, to the nearest whole number? $\newline$ Answer:

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Exponential growth and decay: word problems

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Q. Element

\mathrm{X}

is a radioactive isotope such that every

11

years, its mass decreases by half. Given that the initial mass of a sample of Element

X

5600

grams, how much of the element would remain after

10

years, to the nearest whole number?

\newline

Answer:

Understand the problem: Understand the problem. $\newline$ We need to calculate the remaining mass of a radioactive isotope after $10$ years, knowing that it halves every $11$ years. The initial mass is $5600$ grams.
Determine decay factor per year: Determine the decay factor per year. $\newline$ Since the mass halves every $11$ years, we can use the formula for exponential decay to find the decay factor per year. The decay factor is the $n$ th root of $\frac{1}{2}$ , where $n$ is the number of years for the mass to halve. In this case, $n = 11$ . $\newline$ Decay factor per year = $(\frac{1}{2})^{\frac{1}{11}}$
Calculate remaining mass after $10$ years: Calculate the remaining mass after $10$ years. $\newline$ We apply the decay factor for $10$ years to the initial mass. $\newline$ Remaining mass = Initial mass $\times$ (Decay factor per year) $^{10}$ $\newline$ Remaining mass = $5600 \times \left(\left(\frac{1}{2}\right)^{\frac{1}{11}}\right)^{10}$
Perform the calculation: Perform the calculation. $\newline$ First, calculate the decay factor per year: $\newline$ $(1/2)^{1/11} \approx 0.933033$ $\newline$ Now, apply this factor for $10$ years: $\newline$ Remaining mass $\approx 5600 \times (0.933033)^{10}$ $\newline$ Remaining mass $\approx 5600 \times 0.5^{10/11}$ $\newline$ Remaining mass $\approx 5600 \times 0.933033^{10}$ $\newline$ Remaining mass $\approx 5600 \times 0.5^{10/11}$ $\newline$ Remaining mass $\approx 5600 \times 0.933033^{10}$ $\newline$ Remaining mass $\approx 5600 \times 0.5^{10/11}$ $\newline$ Remaining mass $\approx 5600 \times 0.933033^{10}$