A particle of mass 0.5 MeV/c 2
has kinetic energy 100 eV. Find
its de -Broglie wavelength and group velocity, where c is the
velocity of light.

To find the de Broglie wavelength and group velocity of a particle given its mass and kinetic energy, we'll follow these steps:

### Given Data:
- Mass (\( m \)) = 0.5 MeV/c²
- Kinetic Energy (\( KE \)) = 100 eV
- Speed of light (\( c \)) = \( 3 \times 10^8 \) m/s

### 1. Convert mass from MeV/c² to kg:
The conversion factor is \( 1 \text{ MeV/c}^2 = 1.783 \times 10^{-30} \text{ kg} \).
\[ m = 0.5 \text{ MeV/c}^2 \times 1.783 \times 10^{-30} \text{ kg/MeV/c}^2 = 0.8915 \times 10^{-30} \text{ kg} \]

### 2. Convert kinetic energy from eV to joules:
The conversion factor is \( 1 \text{ eV} = 1.602 \times 10^{-19} \text{ J} \).
\[ KE = 100 \text{ eV} \times 1.602 \times 10^{-19} \text{ J/eV} = 1.602 \times 10^{-17} \text{ J} \]

### 3. Find the particle’s momentum \( p \):
The non-relativistic kinetic energy formula \( KE = \frac{p^2}{2m} \) can be rearranged to find \( p \):
\[ p = \sqrt{2m \cdot KE} \]
\[ p = \sqrt{2 \times 0.8915 \times 10^{-30} \text{ kg} \times 1.602 \times 10^{-17} \text{ J}} \]
\[ p = \sqrt{2 \times 0.8915 \times 1.602 \times 10^{-47}} \]
\[ p = \sqrt{2.854 \times 10^{-47}} \]
\[ p \approx 1.689 \times 10^{-23} \text{ kg} \cdot \text{m/s} \]

### 4. Calculate the de Broglie wavelength \( \lambda \):
The de Broglie wavelength is given by:
\[ \lambda = \frac{h}{p} \]
where \( h \) is Planck’s constant (\( 6.626 \times 10^{-34} \text{ Js} \)).
\[ \lambda = \frac{6.626 \times 10^{-34}}{1.689 \times 10^{-23}} \]
\[ \lambda \approx 3.922 \times 10^{-11} \text{ m} \]
\[ \lambda \approx 0.0392 \text{ nm} \]

### 5. Calculate the group velocity \( v_g \):
The group velocity \( v_g \) for a non-relativistic particle can be approximated by the classical velocity \( v \):
\[ v = \sqrt{\frac{2 \cdot KE}{m}} \]
\[ v = \sqrt{\frac{2 \times 1.602 \times 10^{-17}}{0.8915 \times 10^{-30}}} \]
\[ v = \sqrt{3.594 \times 10^{13}} \]
\[ v \approx 1.895 \times 10^6 \text{ m/s} \]

### Summary:
- **de Broglie wavelength**: \( \lambda \approx 0.0392 \text{ nm} \)
- **Group velocity**: \( v_g \approx 1.895 \times 10^6 \text{ m/s} \)

Do you want details or have any questions?

Here are some related questions:
1. How is the de Broglie wavelength affected by the particle's mass?
2. What would happen to the de Broglie wavelength if the kinetic energy were doubled?
3. How does the de Broglie wavelength relate to the particle's wave-particle duality?
4. What is the significance of the group velocity in wave mechanics?
5. How would the group velocity change if the mass of the particle were halved?
6. What is the difference between phase velocity and group velocity?
7. How is the concept of de Broglie wavelength used in electron microscopy?
8. Can you explain the concept of wave-particle duality with another example?

**Tip:** When dealing with particle physics, always ensure the units are correctly converted to maintain consistency in calculations.

Learn how to calculate the de Broglie wavelength and group velocity of a particle with mass 0.5 MeV/c² and kinetic energy 100 eV. Includes step-by-step calculations and explanations using quantum mechanics principles. Suitable for advanced physics students.

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