- Group velocity
- de Broglie wavelength
(Assume the mass is equal to rest mass of electron)
Answer:-
An electron with kinetic energy of 500 keV is in vacuum. Calculate:
- Group Velocity
- de Broglie Wavelength
Assume mass = rest mass of electron.
Given:
- Kinetic Energy (K.E) = 500 keV = 500 × 10³ × 1.6 × 10⁻¹⁹ = 8 × 10⁻¹⁴ J
- Electron mass (m) = 9.1 × 10⁻³¹ kg
- Planck’s constant (h) = 6.626 × 10⁻³⁴ Js
1. Group Velocity:
Using the classical formula:vg = √(2 × K.E. / m)
Substituting:vg = √(2 × 8 × 10⁻¹⁴ / 9.1 × 10⁻³¹) = √(1.758 × 10¹⁷) ≈ 1.33 × 10⁸ m/s
2. de Broglie Wavelength:
Using the formula:λ = h / (m × v)
Substituting:λ = 6.626 × 10⁻³⁴ / (9.1 × 10⁻³¹ × 1.33 × 10⁸) = 6.626 × 10⁻³⁴ / 1.21 × 10⁻²² ≈ 5.49 × 10⁻¹² m
Final Answers:
- Group Velocity: 1.33 × 10⁸ m/s
- de Broglie Wavelength: 5.49 × 10⁻¹² m