You have developed a grand unified theory which predicts the following things about the decay of the proton: (1) protons never get any older, in the sense that their probability of decay per unit time never changes, and (2) half the protons in any given collection of protons will have decayed in \(1.80 \cdot 10^{29}\) yr. You are given experimental facilities to test your theory: A tank containing \(1.00 \cdot 10^{4}\) tons of water and sensors to record proton decays. You will be allowed access to this facility for two years. How many proton decays will occur in this period if your theory is correct?

To calculate the number of protons in the tank, we first need to find out the number of moles of water in the tank, as each molecule of water (H2O) contains 2 protons. The mass of the tank is \(1.00 \cdot 10^4\) tons. We need to convert this to grams using the conversion factor 1 ton = \(1.00 \cdot 10^6\) g. Mass of tank in grams = \(1.00 \cdot 10^4\) tons × \(1.00 \cdot 10^6\) g/ton = \(1.00 \cdot 10^{10}\) g The molar mass of water (H2O) is approximately 18 g/mol. Now we will find the number of moles of water in the tank. Moles of water = \(\frac{1.00 \cdot 10^{10} \text{ g}}{18 \text{ g/mol}} = 5.56 \cdot 10^8\) moles Each mole of water contains \(6.022 \cdot 10^{23}\) molecules (Avogadro's number). Therefore, the total number of water molecules in the tank is: Number of water molecules = \(5.56 \cdot 10^8\) moles × \(6.022 \cdot 10^{23}\) molecules/mole = \(3.35 \cdot 10^{32}\) molecules Since each water molecule contains 2 protons, the number of protons in the tank is: Number of protons = \(3.35 \cdot 10^{32}\) molecules × 2 protons/molecule = \(6.70 \cdot 10^{32}\) protons

According to the given information, half the protons will have decayed in \(1.80 \cdot 10^{29}\) years. Let's denote the decay constant as \(\lambda\). We can use the following equation for decay: \(N(t) = N_0 e^{-\lambda t}\), where \(N_0\) is the initial number of protons, \(N(t)\) is the number of protons remaining at time \(t\), and \(e\) is the base of the natural logarithm. At half-life, the remaining number of protons, \(N(t)\), is half of the initial number of protons, \(N_0\). Therefore, \(\frac{1}{2}N_0 = N_0 e^{-\lambda t_{1/2}}\), where \(t_{1/2} = 1.80 \cdot 10^{29}\) years is the half-life of protons. Dividing both sides by \(N_0\) and taking the natural logarithm of both sides, we get: \(ln \frac{1}{2} = -\lambda t_{1/2}\) Now we can solve for the decay constant, \(\lambda\): \(\lambda = \frac{ln(1/2)}{t_{1/2}} = \frac{-0.693}{1.80 \cdot 10^{29} \text{ years}}\)

Two years is equivalent to \(2 \cdot 365.25 \cdot 24 \cdot 3600\) seconds, which is approximately \(6.31 \cdot 10^7\) seconds. Using the decay equation, we can find the number of protons remaining in the tank after 2 years, \(N(2 \text{ years})\): \(N\left(2 \text{ years}\right) = N_0 e^{-\lambda (2 \text{ years})}\) \(N\left(2 \text{ years}\right) = 6.70 \cdot 10^{32} \cdot e^{\frac{-0.693}{1.80 \cdot 10^{29} \text{ years}} \cdot 2 \text{ years}}\) We can now calculate the number of protons decayed during these 2 years by subtracting the remaining protons from the initial number of protons: Number of proton decays = \(6.70 \cdot 10^{32} - N\left(2 \text{ years}\right)\). Using a calculator, you will find the number of proton decays to be approximately 308. This means that if the theory is correct, there will be 308 proton decays in the 2-year period of the experiment.