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A proton or a neutron can sometimes "violate" conservation of energy by emitting and then reabsorbing a pi meson, which has a mass of 135 MeV/c \(^{2}\). This is possible as long as the pi meson is reabsorbed within a short enough time \(\Delta t\) consistent with the uncertainty principle, (a) Consider \(\mathrm{p} \rightarrow \mathrm{p}+\pi .\) By what amount \(\Delta E\) is energy conservation violated? (Ignore any kinetic energies) (b) For how long a time \(\Delta t\) can the pi meson exist? (c) Assuming that the pi meson travels at very nearly the speed of light, how far from the proton can it go? (This procedure, as we discuss in Chapter \(12,\) gives us an estimate of the range of the nuclear force, because protons and neutrons are held together in the nucleus by exchanging pi mesons.)

Step by step solution

01

Determine \( \Delta E \) Violation

When a proton emits a pi meson, the energy conservation is violated by an amount \( \Delta E = m_{\pi} c^{2} \), where \( m_{\pi} \) is the mass of the pi meson (135 MeV/c^2). Thus, \( \Delta E = 135 \ \text{MeV} \).

02

Apply the Uncertainty Principle

According to the uncertainty principle, \( \Delta E \Delta t \approx \hbar \), where \( \hbar \approx 6.58 \times 10^{-16} \ \text{eV} \cdot\text{s} \). Rearrange to find \( \Delta t \). \[ \Delta t = \frac{\hbar}{\Delta E} = \frac{6.58 \times 10^{-16} \ \text{eV} \cdot\text{s}}{135 \times 10^6 \ \text{eV}} \approx 4.87 \times 10^{-24} \ \text{s} \].

03

Determine the Distance Traveled

If the pi meson travels very nearly at the speed of light \( c \approx 3.00 \times 10^{8} \ \text{m/s} \, use the relationship \ x = c \Delta t \). Hence, the distance \[ x \approx 3.00 \times 10^{8} \ \text{m/s} \times 4.87 \times 10^{-24} \ \text{s} \approx 1.46 \times 10^{-15} \ \text{m} \].

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

energy conservation violation

In physics, the law of energy conservation states that energy cannot be created or destroyed, only transformed from one form to another. However, certain quantum phenomena appear to violate this principle temporarily. When a proton or neutron emits a pi meson (\text{蟺 meson}), it seems like energy conservation is violated.
But, this 'violation' follows the rules set by the uncertainty principle, which allows such occurrences within very short periods. Here, the mass of the pi meson (\text{135 MeV/c\(^{2}\)}) translates to an energy violation amount, \text{螖E} = 135 MeV.
This doesn't mean energy is truly lost or created, but rather that it's borrowed and quickly repaid, keeping the overall system balanced.

pi meson emission

Pi mesons are subatomic particles involved in the strong nuclear force. Protons and neutrons can emit these particles, which are quickly reabsorbed. This emission and reabsorption can be understood using the uncertainty principle.
According to the principle, \text{螖E} \text{螖t} \text { = } \text{鈩弣, where \text{鈩弣 is roughly 6.58 脳 10\(^{-16}\) \text{eV} 路 \text{s}. In our calculation, we found \text {螖t} to be roughly 4.87 脳 10\(^{-24}\) \text {s}.
This extremely short duration makes it possible to 'violate' energy conservation without any long-term effects. When a proton emits a pi meson and reabsorbs it within this timeframe, the energy discrepancy doesn't disrupt the overall energy balance.

nuclear force range

The range of the nuclear force is critical for the stability of atomic nuclei. Pi meson emission is key to understanding this range. The pi meson, traveling nearly at the speed of light, determines how far it can go before being reabsorbed.
Using the speed of light (c 鈮� 3.00 脳 10\(^{8}\) \text {m/s} and our calculated time (4.87 脳 10\(^{-24}\) \text {s}), we estimate the distance to be about 1.46 脳 10\(^{-15}\) \text{m}.
This distance represents the effective range of the nuclear force, showing how protons and neutrons interact within the nucleus. Exchanging pi mesons over such short distances ensures the strong nuclear force holds particles together.