Argon- 41 is used to measure the rate of gas flow. It has a decay constant of \(6.3 \times 10^{-3} \mathrm{~min}^{-1}\). (a) What is its half-life? (b) How long will it take before only \(1.00 \%\) of the original amount of Ar- 41 is left?

To find the half-life of Argon-41, we will use the equation \(t_{1/2} = \frac{\ln 2}{\lambda}\). We are given the decay constant \(\lambda = 6.3 \times 10^{-3} \mathrm{~min}^{-1}\). Plugging this into the equation, we get: \(t_{1/2} = \frac{\ln 2}{6.3 \times 10^{-3} \mathrm{~min}^{-1}}\) Calculate the value of \(t_{1/2}\).

We now want to find the time \(t\) when only 1% of the original amount of Argon-41 is left. We can use the decay formula \(N_t = N_0 e^{-\lambda t}\). Since we are only concerned with the percentage of remaining Argon-41, we can set \(N_0 = 100\) and \(N_t = 1\). So, the equation becomes: \(1 = 100 e^{-6.3 \times 10^{-3} \mathrm{~min}^{-1} t}\) To solve for \(t\), we can first divide both sides by 100: \(\frac{1}{100} = e^{-6.3 \times 10^{-3} \mathrm{~min}^{-1} t}\) Next, take the natural logarithm of both sides: \(\ln(\frac{1}{100}) = -6.3 \times 10^{-3} \mathrm{~min}^{-1} t\) Now, divide both sides by the decay constant to isolate \(t\): \(t = \frac{\ln(\frac{1}{100})}{-6.3 \times 10^{-3} \mathrm{~min}^{-1}}\) Finally, calculate the value of \(t\).