Question

An electron, a proton and an alpha particle are moving with same kinetic energy. If $ \lambda_1, \lambda_2, \lambda_3 $ are the de Broglie wavelengths of the electron, proton and alpha particle respectively, then

• A. \( \lambda_1 = \lambda_2 = \lambda_3 \)
• B. \( \lambda_1>\lambda_2>\lambda_3 \)
• C. \( \lambda_1<\lambda_2<\lambda_3 \)
• D. \( \lambda_1 = \lambda_2 \neq \lambda_3 \)

Hint:

For particles with the same kinetic energy, the de Broglie wavelength is inversely proportional to the square root of the mass.

Answer 1

The Correct Option is B

The de Broglie wavelength of a particle is given by: \[ \lambda = \frac{h}{\sqrt{2mK}} \] where \( h \) is Planck's constant, \( m \) is the mass of the particle, and \( K \) is its kinetic energy. Since the kinetic energy is the same for all particles, the particle with the smallest mass will have the largest wavelength. Therefore, the order of wavelengths is \( \lambda_1>\lambda_2>\lambda_3 \), where \( \lambda_1 \) is the wavelength of the electron, the lightest particle.