1. **Problem statement:** We have a radioactive isotope Technetium-99m with a half-life of 6 hours. The initial mass is 2 grams. We want to find: a) The mass remaining after $t$ hours. b) The time it takes for the mass to decay to 1.5 grams. --- 2. **Formula used:** The decay of a radioactive substance follows the exponential decay formula: $$m(t) = m_0 \times \left(\frac{1}{2}\right)^{\frac{t}{T}}$$ where: - $m(t)$ is the mass at time $t$, - $m_0$ is the initial mass, - $T$ is the half-life, - $t$ is the elapsed time. --- 3. **Part (a): Find $m(t)$ given $m_0=2$ grams and half-life $T=6$ hours.** The formula becomes: $$m(t) = 2 \times \left(\frac{1}{2}\right)^{\frac{t}{6}}$$ This expression gives the mass remaining after $t$ hours. --- 4. **Part (b): Find $t$ when $m(t) = 1.5$ grams.** Start with the formula: $$1.5 = 2 \times \left(\frac{1}{2}\right)^{\frac{t}{6}}$$ Divide both sides by 2: $$\frac{1.5}{2} = \left(\frac{1}{2}\right)^{\frac{t}{6}}$$ Show cancellation: $$\frac{1.5}{\cancel{2}} = \left(\frac{1}{2}\right)^{\frac{t}{6}}$$ Simplify: $$0.75 = \left(\frac{1}{2}\right)^{\frac{t}{6}}$$ Take natural logarithm on both sides: $$\ln(0.75) = \ln\left(\left(\frac{1}{2}\right)^{\frac{t}{6}}\right) = \frac{t}{6} \ln\left(\frac{1}{2}\right)$$ Solve for $t$: $$t = 6 \times \frac{\ln(0.75)}{\ln(\frac{1}{2})}$$ Calculate the values: $$\ln(0.75) \approx -0.28768, \quad \ln(\frac{1}{2}) = \ln(0.5) \approx -0.69315$$ So: $$t = 6 \times \frac{-0.28768}{-0.69315} = 6 \times 0.414 = 2.484 \text{ hours}$$ --- **Final answers:** - a) The mass remaining after $t$ hours is: $$m(t) = 2 \times \left(\frac{1}{2}\right)^{\frac{t}{6}}$$ - b) The time to decay to 1.5 grams is approximately: $$t \approx 2.48 \text{ hours}$$