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 larger for lighter particles when moving at the same speed. Always compare the masses to determine the correct wavelength order.

Answer 1

The Correct Option is B

The de Broglie wavelength of a particle is given by: \[ \lambda = \frac{h}{m v} \] where: 
\( h \) is Planck’s constant,
\( m \) is the mass of the particle,
\( v \) is the velocity of the particle.
Since all three particles (electron, proton, and deuteron) have the same speed, the wavelength is inversely proportional to their masses: \[ \lambda \propto \frac{1}{m} \] The masses of the particles are:
\( m_e \) (electron) is the smallest,
\( m_p \) (proton) is larger,
\( m_d \) (deuteron) is the largest.
Thus, their wavelengths follow the relation: \[ \lambda_e > \lambda_p > \lambda_d \] Hence, the correct answer is (2).