Question

The Louis de Broglie wavelength of a particle having kinetic energy E is:

• A. \( \lambda = \frac{h}{\sqrt{2mE}} \)
• B. \( \lambda = \frac{h}{\sqrt{mE}} \)
• C. \( \lambda = \frac{\sqrt{2mE}}{h} \)
• D. \( \lambda = \frac{\sqrt{mE}}{h} \)

Hint:

The de Broglie wavelength is inversely related to the square root of the particle’s kinetic energy and its mass.

Answer 1

The Correct Option is A

The de Broglie wavelength \( \lambda \) of a particle with kinetic energy \( E \) is given by the relation: \[ \lambda = \frac{h}{\sqrt{2mE}}, \] where \( h \) is Planck’s constant, \( m \) is the mass of the particle, and \( E \) is its kinetic energy.
Correct Answer: (A) \( \lambda = \frac{h}{\sqrt{2mE}} \).