Problem 33 - Physics for Scientists and Engineers with Modern Physics

Question

The mean life of the \(\Sigma^{0}\) particle is \(7 \times 10^{-20} \mathrm{~s}\). What is the uncertainty in its rest energy? Express your answer in MeV.

Step-by-Step Solution

Answer

The uncertainty in the rest energy of the \( \Sigma^{0} \) particle is approximately \( 4.70 \mathrm{~MeV} \).

Step 1: Identify Given Values

We are given that the mean life of the \( \Sigma^{0} \) particle is \( 7 \times 10^{-20} \mathrm{~s} \). This is the lifetime \( \tau \) of the particle.

Step 2: Use the Energy-Time Uncertainty Principle

According to the energy-time uncertainty principle, the uncertainty in energy \( \Delta E \) is given by \( \Delta E \cdot \Delta t \geq \frac{\hbar}{2} \), where \( \hbar \) is the reduced Planck's constant \( \hbar = 1.0545718 \times 10^{-34} \mathrm{~Js} \). Here, \( \Delta t \) can be approximated by the mean life \( \tau \).

Step 3: Calculate the Uncertainty in Energy

First, calculate \( \Delta E \) using \( \Delta E \approx \frac{\hbar}{2\tau} \). Substitute \( \hbar = 1.0545718 \times 10^{-34} \mathrm{~Js} \) and \( \tau = 7 \times 10^{-20} \mathrm{~s} \):\[ \Delta E = \frac{1.0545718 \times 10^{-34}}{2 \times 7 \times 10^{-20}} \approx 7.5327 \times 10^{-16} \mathrm{~J} \]

Step 4: Convert Joules to MeV

To convert the energy from Joules to MeV, use the conversion factor \( 1 \mathrm{~J} = 6.242 \times 10^{12} \mathrm{~MeV} \). Multiply \( 7.5327 \times 10^{-16} \mathrm{~J} \) by \( 6.242 \times 10^{12} \mathrm{~MeV/J} \):\[ \Delta E \approx 7.5327 \times 10^{-16} \times 6.242 \times 10^{12} \approx 4.70 \mathrm{~MeV} \]

Step 5: Conclusion

The uncertainty in the rest energy of the \( \Sigma^{0} \) particle is approximately \( 4.70 \mathrm{~MeV} \).

Key Concepts

Lifetime of particles

In the realm of particle physics, the concept of a particle's lifetime is a vital aspect. The lifetime of a particle is the average time a particle exists before decaying into other particles. It's crucial because it provides insight into the stability and nature of the particle.
For instance, the lifetime of the \( \Sigma^{0} \) particle is given as \( 7 \times 10^{-20} \) seconds. This is an incredibly short time, illustrating that \( \Sigma^{0} \) is a highly unstable particle.
Understanding the lifetime helps physicists predict how particles behave under different circumstances, which is essential for experiments and theoretical physics.
Remember that lifetime is directly tied to the uncertainty in a particle's energy, due to the Energy-Time Uncertainty Principle. The shorter the lifetime, the greater the energy uncertainty. This relation plays a crucial role in predicting the particle’s behavior in quantum mechanics.

Reduced Planck's constant

Planck's constant is a fundamental constant in physics that indicates the scale of quantum effects. The reduced Planck's constant, \( \hbar \), is simply the Planck constant divided by \( 2\pi \).
The value of \( \hbar \) is \( 1.0545718 \times 10^{-34} \mathrm{~Js} \), and it appears extensively in quantum mechanics equations.

• It is used to quantify the scales at which quantum mechanical effects become significant.
• It plays a crucial role in the energy-time uncertainty relation \( \Delta E \cdot \Delta t \geq \frac{\hbar}{2} \), where \( \Delta E \) is the uncertainty in energy and \( \Delta t \) is the uncertainty in time.

The reduced Planck's constant sets the limit for the precision of these measurements. Without it, modeling quantum phenomena accurately would be nearly impossible. Therefore, understanding \( \hbar \) is crucial for deciphering how particles like \( \Sigma^{0} \) behave on a quantum level.

Conversion of energy units

When dealing with particle physics problems, energy is often measured in different units. The most common units include Joules (J) and Mega-electronvolts (MeV).
The unit conversion is necessary because Joules are a standard SI unit for energy, while electronvolts are more practical in high-energy physics.
To convert from Joules to MeV, use the conversion factor: \( 1 \mathrm{~J} = 6.242 \times 10^{12} \mathrm{~MeV} \).

• Multiply the energy value in Joules by the conversion factor to obtain energy in MeV.
• This conversion helps make calculations and interpretations more intuitive in the context of particle reactions and decays.

For the \( \Sigma^{0} \) particle, the calculated energy in Joules was \( 7.5327 \times 10^{-16} \mathrm{~J} \) and converting it gives us approximately \( 4.70 \mathrm{~MeV} \).
This result is significant because MeV is a manageable unit when discussing particle energies, making the results more straightforward to interpret and compare across different contexts or experiments.