Understanding the Concepts
* De Broglie Wavelength: Louis de Broglie proposed that particles, like electrons, can exhibit wave-like behavior. The wavelength associated with a particle is given by:
λ = h / p
where:
* λ is the de Broglie wavelength
* h is Planck's constant (6.626 x 10⁻³⁴ J·s)
* p is the momentum of the particle (p = mv, where m is mass and v is velocity)
Calculations
1. Momentum:
* Mass of an electron (m) = 9.11 x 10⁻³¹ kg
* Velocity of the electron (v) = 9.38 x 10⁶ m/s
* Momentum (p) = (9.11 x 10⁻³¹ kg) * (9.38 x 10⁶ m/s) = 8.54 x 10⁻²⁴ kg·m/s
2. De Broglie Wavelength:
* λ = (6.626 x 10⁻³⁴ J·s) / (8.54 x 10⁻²⁴ kg·m/s)
* λ ≈ 7.76 x 10⁻¹¹ m
Answer
The characteristic wavelength of the electron accelerated to a speed of 9.38 x 10⁶ m/s is approximately 7.76 x 10⁻¹¹ meters. This wavelength falls in the X-ray region of the electromagnetic spectrum.
Important Note: The question specifically asks for the characteristic wavelength based on the electron's speed. If you had the potential difference the electron was accelerated through, you could also calculate the wavelength using the relationship between potential energy and kinetic energy.
