The amount of Cu-67 in the blood of a patient may be modelled by the following equation. m(t)=m_(0)((1)/(2))^((t)/(h)). where m(0)=m_(0) is the initial amount of Cu-67, that is, the amount present at the time of injection. Here t represents time measured in days from the time of injection and h=2.6 days is the half-life. Sketch the graph of m(t) over a period of just over five half-lives from the time of injection (at time t=0). Assume m_(0)=8mg. Find the ratio (m(t))/(m_(0)) predicted by the model 10.4 days after injection. This is the fraction of the Cu-67 content remaining. Plot the corresponding point on your graph. When the mass of Cu-67 falls below 0.2 mg the radiation can no longer be detected. How long after the injection must the patient wait until the radiation can no longer be detected?

Question

The amount of Cu-67 in the blood of a patient may be modelled by the following equation.    m(t)=m_(0)((1)/(2))^((t)/(h)). where m(0)=m_(0) is the initial amount of Cu-67, that is, the amount present at the time of injection. Here t represents time measured in days from the time of injection and h=2.6 days is the half-life. Sketch the graph of m(t) over a period of just over five half-lives from the time of injection (at time t=0). Assume m_(0)=8mg.  Find the ratio (m(t))/(m_(0)) predicted by the model 10.4 days after injection. This is the fraction of the Cu-67 content remaining. Plot the corresponding point on your graph. When the mass of Cu-67 falls below 0.2 mg the radiation can no longer be detected. How long after the injection must the patient wait until the radiation can no longer be detected?

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Decay Function Explanation: To sketch the graph of m(t), we need to understand the decay function given by m(t) = m_0 * (1/2)^(t/h). Since the half-life h is 2.6 days, the function will halve the initial amount m_0 every 2.6 days. We will plot the graph starting from t=0 up to just over five half-lives, which is just over 5 * 2.6 days.Calculate Remaining Amount: To find the ratio (m(t))/m_0 after 10.4 days, we substitute t with 10.4 days and h with 2.6 days into the decay function. The initial amount m_0 is given as 8 mg. m(t) = m_0 * (1/2)^(t/h) = 8 * (1/2)^(10.4/2.6)Plotting Graph Point: Calculate the exponent: 10.4 / 2.6 = 4. This means that 10.4 days is exactly 4 half-lives.Inequality Setup: Now we can calculate the remaining amount of Cu-67 after 10.4 days: m(t) = 8 * (1/2)^4 = 8 * (1/16) = 8 / 16 = 0.5 mgIsolate Exponential Term: The ratio (m(t))/m_0 is then 0.5 mg / 8 mg = 1/16.Logarithm Property: Plot the point (10.4, 0.5) on the graph to represent the amount of Cu-67 remaining after 10.4 days.Solve for t: To find out when the mass of Cu-67 falls below 0.2 mg, we set up the inequality m(t) < 0.2 mg and solve for t. 0.2 > m_0 * (1/2)^(t/h) 0.2 > 8 * (1/2)^(t/2.6)Calculate Right Side: Divide both sides by 8 to isolate the exponential term: 0.2 / 8 > (1/2)^(t/2.6) 0.025 > (1/2)^(t/2.6)Final Result: To solve for t, we take the logarithm of both sides. We use the property that log(a^b) = b * log(a). log(0.025) > (t/2.6) * log(1/2)Now we solve for t: t/2.6 > log(0.025) / log(1/2) t > 2.6 * (log(0.025) / log(1/2))Calculate the right side of the inequality using a calculator: t > 2.6 * (log(0.025) / log(1/2)) ≈ 2.6 * (-1.60206 / -0.30103) ≈ 2.6 * 5.32048 ≈ 13.83325The patient must wait approximately 13.83 days after the injection for the radiation to no longer be detectable, as this is when the mass of Cu-67 falls below 0.2 mg.

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