This exercise uses the radioactive decay model. $\newline$ The half-life of strontium $-90$ is $29$ years. How long (in yr) will it take a $70$ -milligram sample to decay to a mass of $56$ mg? (Round your answer to the nearest whole number.) $\newline$ yr

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Q. This exercise uses the radioactive decay model.

\newline

The half-life of strontium

-90

29

years. How long (in yr) will it take a

70

-milligram sample to decay to a mass of

56

mg? (Round your answer to the nearest whole number.)

\newline

Understand half-life concept: Understand the half-life concept and set up the decay formula. $\newline$ The half-life of a substance is the time it takes for half of the substance to decay. For strontium $-90$ , the half-life is $29$ years. The decay formula for a substance with a half-life (h) is given by: $\newline$ $y = a \left(\frac{1}{2}\right)^{\frac{x}{h}}$ $\newline$ where: $\newline$ - $y$ is the final amount of the substance, $\newline$ - $a$ is the initial amount of the substance, $\newline$ - $x$ is the time elapsed, $\newline$ - $h$ is the half-life of the substance.
Insert known values: Insert the known values into the decay formula. $\newline$ We know that the initial amount $a$ is $70$ mg, the final amount $y$ we want is $56$ mg, and the half-life $h$ is $29$ years. We want to find $x$ , the time it takes to decay from $70$ mg to $56$ mg. So we set up the equation: $\newline$ $56 = 70 \left(\frac{1}{2}\right)^{\frac{x}{29}}$
Solve for x: Solve for $x$ . $\newline$ First, divide both sides of the equation by $70$ to isolate the exponential part: $\newline$ $\frac{56}{70} = \left(\frac{1}{2}\right)^{\frac{x}{29}}$ $\newline$ Simplify the left side of the equation: $\newline$ $\frac{4}{5} = \left(\frac{1}{2}\right)^{\frac{x}{29}}$
Take logarithm: Take the logarithm of both sides to solve for $x$ . $\newline$ We can use the natural logarithm (ln) to help solve for $x$ : $\newline$ $\ln\left(\frac{4}{5}\right) = \ln\left(\left(\frac{1}{2}\right)^{\frac{x}{29}}\right)$ $\newline$ Using the power rule of logarithms, we can move the exponent in front of the ln: $\newline$ $\ln\left(\frac{4}{5}\right) = \frac{x}{29} \cdot \ln\left(\frac{1}{2}\right)$
Isolate and solve: Isolate $x$ and solve. $\newline$ Divide both sides by $\ln\left(\frac{1}{2}\right)$ to get $x$ by itself: $\newline$ $x = \frac{\ln\left(\frac{4}{5}\right)}{\ln\left(\frac{1}{2}\right)} \cdot 29$ $\newline$ Now we can use a calculator to find the value of $x$ : $\newline$ $x \approx \frac{\ln(0.8)}{\ln(0.5)} \cdot 29$ $\newline$ $x \approx \frac{-0.22314}{-0.69315} \cdot 29$ $\newline$ $x \approx 0.32193 \cdot 29$ $\newline$ $x \approx 9.336$ $\newline$ Round the answer to the nearest whole number: $\newline$ $x \approx 9$