The energy-time uncertainty principle allows for the creation of virtual particles, that appear from a vacuum for a very brief period of time $\Delta t,\( then disappear again. This can happen as long as \)\Delta E \Delta t=\hbar / 2,\( where \)\Delta E$ is the rest energy of the particle. (a) How long could an electron created from the vacuum exist according to the uncertainty principle? (b) How long could a shot put with a mass of \(7 \mathrm{kg}\) created from the vacuum exist according to the uncertainty principle?

To find the time duration for an electron, we must first determine its rest energy. The equation for rest energy is \(E_0 = mc^2\). Here, m denotes the mass of an electron and c is the speed of light. We also know from the problem statement that \(\Delta E \Delta t = \frac{\hbar}{2}\). By plugging the rest energy, we can solve for the time duration \(\Delta t\). The mass of an electron is \(9.11 \times 10^{-31}\) kg and the speed of light is \(3.0 \times 10^8\) m/s. First, we will calculate the rest energy of the electron: \(E_0 = (9.11 \times 10^{-31} \,\text{kg})(3.0 \times 10^8\, \text{m/s})^2 \approx 8.19 \times 10^{-14} \,\text{J}\) Now, we will solve for the time duration, \(\Delta t\), using the energy-time uncertainty principle: \(\Delta E \Delta t = \frac{\hbar}{2}\) \(\Delta t = \frac{\hbar}{2\Delta E}\) Here, \(\hbar\) is the reduced Planck constant, which has a value of approximately \(1.05 \times 10^{-34} \,\text{Js}\). Now, plug in the values: \(\Delta t = \frac{1.05 \times 10^{-34} \,\text{Js}}{2(8.19 \times 10^{-14} \,\text{J})} \approx 6.40 \times 10^{-22}\, \text{s}\) So, an electron created from the vacuum could exist for approximately \(6.40 \times 10^{-22}\) seconds according to the uncertainty principle.

To find the time duration for a shot put with a mass of 7 kg, we must first determine its rest energy, again using the equation \(E_0 = mc^2\). We also use the same formula for the energy-time uncertainty principle as in part (a) to solve for the time duration. First, calculate the rest energy of the shot put: \(E_0 = (7 \,\text{kg})(3.0 \times 10^8\, \text{m/s})^2 \approx 6.30 \times 10^{17} \,\text{J}\) Now, we will solve for the time duration, \(\Delta t\), using the energy-time uncertainty principle: \(\Delta E \Delta t = \frac{\hbar}{2}\) \(\Delta t = \frac{\hbar}{2\Delta E}\) Plug in the values: \(\Delta t = \frac{1.05 \times 10^{-34} \,\text{Js}}{2(6.30 \times 10^{17} \,\text{J})} \approx 8.33 \times 10^{-53}\, \text{s}\) So, a shot put with a mass of 7 kg created from the vacuum could exist for approximately \(8.33 \times 10^{-53}\) seconds according to the uncertainty principle.