Question

The de Broglie wavelength associated with the electrons accelerated by a potential of 81 V is lying in the region of electromagnetic waves:

• A. Ultraviolet rays
• B. Infrared rays
• C. Microwaves
• D. X-rays
• E. \( \gamma \)-rays

Hint:

The de Broglie wavelength of a particle depends on its momentum. For high-energy electrons, such as those accelerated by 81 V, the wavelength lies in the X-ray region of the electromagnetic spectrum.

Answer 1

The Correct Option is D

The de Broglie wavelength \( \lambda \) associated with a particle (in this case, an electron) is given by the formula: \[ \lambda = \frac{h}{p} \] where: - \( h \) is Planck's constant (\( 6.626 \times 10^{-34} \, {J·s} \)),
- \( p \) is the momentum of the particle.
The momentum \( p \) of an electron accelerated by a potential \( V \) is: \[ p = \sqrt{2 m_e e V} \] where: - \( m_e \) is the mass of the electron (\( 9.11 \times 10^{-31} \, {kg} \)),
- \( e \) is the charge of the electron (\( 1.6 \times 10^{-19} \, {C} \)),
- \( V \) is the potential difference (in this case, 81 V).
Substituting these values into the formula for \( p \): \[ p = \sqrt{2 \times 9.11 \times 10^{-31} \times 1.6 \times 10^{-19} \times 81} \] Now, the de Broglie wavelength \( \lambda \) can be calculated using the above formula. For this case, the wavelength \( \lambda \) will be in the range of X-rays.
Therefore, the correct answer is: \[ \boxed{{D) X-rays}} \]