Question

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

Hint:

The de Broglie wavelength is inversely proportional to the mass of the particle. Heavier particles have smaller wavelengths.

Answer 1

The Correct Option is A

The de Broglie wavelength \( \lambda \) of a particle is given by:

\[ \lambda = \frac{h}{p} \]

where \( h \) is Planck's constant and \( p \) is the momentum of the particle. Since the momentum \( p = mv \), and all particles are moving with the same speed, the mass of the particle determines the wavelength.

Since the mass of the electron is the smallest, followed by the proton and deuteron, the wavelength of the electron will be the largest, followed by the proton and then the deuteron.

Thus, the correct relation is \( \lambda_e > \lambda_p > \lambda_d \).