Question

What is the de Broglie wavelength of the particle having kinetic energy of $ 2E $?

• A. \( \frac{h}{\sqrt{2mE}} \)
• B. \( \frac{h}{\sqrt{2m2E}} \)
• C. \( \frac{h}{2\sqrt{mE}} \)
• D. \( \frac{h}{\sqrt{mE}} \)

Hint:

The de Broglie wavelength depends on the momentum of the particle, and kinetic energy is directly related to the square of the particle's velocity, which affects the wavelength.

Answer 1

The Correct Option is B

The de Broglie wavelength \( \lambda \) is given by the formula: \[ \lambda = \frac{h}{p} \] Where: - \( h \) is Planck's constant. - \( p \) is the momentum of the particle. Momentum can be expressed as: \[ p = \sqrt{2mE} \] Where: - \( m \) is the mass of the particle. - \( E \) is the kinetic energy. Substituting this into the de Broglie equation: \[ \lambda = \frac{h}{\sqrt{2mE}} \] But here we are given that the kinetic energy is \( 2E \), so the new momentum becomes: \[ \lambda = \frac{h}{\sqrt{2m2E}} = \frac{h}{\sqrt{4mE}} = \frac{h}{2\sqrt{mE}} \] Hence, the correct answer is \( \frac{h}{2\sqrt{mE}} \).