Question

this exercise uses the radioactive decay model. the half-life of cesium-137 is 30 years. suppose we have a 11-g sample. (a) find a function $m(t) = m_02^{-t/h}$ that models the mass remaining after $t$ years. $m(t) = \square$ (b) how much of the sample will remain after 61 years? (round your answer to one decimal place.) $\square$ g (c) after how many years will only 2 g of the sample remain? (round your answer to the nearest whole number.) $\square$ yr

Explanation:

Part (a)

Step1: Identify \( m_0 \) and \( h \)

The initial mass \( m_0 \) is 11 g (given sample mass), and the half - life \( h \) is 30 years (given half - life of cesium - 137).

Step2: Substitute into the formula

The formula for radioactive decay is \( m(t)=m_02^{-t/h} \). Substituting \( m_0 = 11 \) and \( h = 30 \) into the formula, we get \( m(t)=11\times2^{-t/30} \).

Step1: Use the function from part (a)

We have \( m(t)=11\times2^{-t/30} \), and we need to find \( m(61) \). Substitute \( t = 61 \) into the function: \( m(61)=11\times2^{-61/30} \).

Step2: Calculate the exponent

First, calculate \( \frac{61}{30}\approx2.0333 \). Then \( 2^{-2.0333}=\frac{1}{2^{2.0333}} \). We know that \( 2^{2}=4 \) and \( 2^{0.0333}\approx1.024 \), so \( 2^{2.0333}\approx4\times1.024 = 4.096 \), and \( \frac{1}{4.096}\approx0.244 \).

Step3: Multiply by 11

\( m(61)=11\times0.244\approx2.7 \) (rounded to one decimal place).

Step1: Set up the equation

We know that \( m(t) = 2 \), and \( m(t)=11\times2^{-t/30} \). So we set up the equation \( 2 = 11\times2^{-t/30} \).

Step2: Solve for the exponent term

Divide both sides of the equation by 11: \( \frac{2}{11}=2^{-t/30} \).

Step3: Take the logarithm of both sides

Take the logarithm (base 2) of both sides. Recall that \( \log_b(a^c)=c\log_b(a) \) and \( \log_b(b)=1 \). So \( \log_2(\frac{2}{11})=\log_2(2^{-t/30}) \). The right - hand side simplifies to \( -\frac{t}{30} \) (since \( \log_2(2^x)=x \)). And \( \log_2(\frac{2}{11})=\log_2(2)-\log_2(11)=1 - \log_2(11) \). We know that \( \log_2(11)\approx3.459 \), so \( \log_2(\frac{2}{11})\approx1 - 3.459=-2.459 \).

Step4: Solve for \( t \)

We have \( - 2.459=-\frac{t}{30} \). Multiply both sides by - 30: \( t = 2.459\times30\approx73.77\approx74 \) (rounded to the nearest whole number).

Answer:

\( m(t)=11\cdot2^{-t/30} \)

Part (b)