Question

Let \( \lambda_e \), \( \lambda_p \), and \( \lambda_d \) be the wavelengths associated with an electron, a proton, and a deuteron, all moving with the same speed. Then the correct relation between them is:

• A. \( \lambda_d>\lambda_p>\lambda_e \)
• B. \( \lambda_e>\lambda_p>\lambda_d \)
• C. \( \lambda_p>\lambda_e>\lambda_d \)
• D. \( \lambda_e = \lambda_p = \lambda_d \)

Hint:

The de Broglie wavelength is inversely proportional to the mass of the particle. The more massive the particle, the smaller its wavelength.

Answer 1

The Correct Option is A

The wavelength \( \lambda \) of a particle is related to its momentum \( p \) by the de Broglie relation: \[ \lambda = \frac{h}{p} \] where \( h \) is Planck's constant and \( p \) is the momentum of the particle. Since all particles are moving with the same speed, the momentum of a particle is given by: \[ p = mv \] where \( m \) is the mass of the particle and \( v \) is its speed. Therefore, the wavelength is inversely proportional to the mass of the particle. Since a deuteron has the largest mass, followed by a proton and then an electron, the deuteron will have the largest wavelength, the proton a smaller wavelength, and the electron will have the smallest wavelength. Thus, the correct order is \( \lambda_d>\lambda_p>\lambda_e \).