Question

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?

Short answer

Answer: (a) An electron can exist as a virtual particle created from the vacuum for approximately \(6.40 \times 10^{-22}\) seconds. (b) A shot put with a mass of 7 kg can exist as a virtual particle created from the vacuum for approximately \(8.33 \times 10^{-53}\) seconds.

Step-by-step solution

Part (a): Time duration for an electron

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.

Part (b): Time duration for a shot put

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.