Question

Electrons and protons are accelerated by the same potential difference. The ratio of their de Broglie wavelengths is:

• A. \( \frac{m_e}{m_p} \)
• B. \( \frac{m_p}{m_e} \)
• C. 1
• D. \( \sqrt{\frac{m_p}{m_e}} \)

Hint:

In n-type semiconductors, electrons are the majority charge carriers, and holes are the minority charge carriers.

Answer 1

The Correct Option is D

The de Broglie wavelength \( \lambda \) of a particle is given by the formula:
\[ \lambda = \frac{h}{p} \] Where:
- \( h \) is Planck's constant,
- \( p \) is the momentum of the particle.
The momentum \( p \) of a particle is given by:
\[ p = mv \] Where:
- \( m \) is the mass of the particle,
- \( v \) is the velocity of the particle.
Since both the electron and the proton are accelerated by the same potential difference \( V \), their kinetic energy is:
\[ K.E. = \frac{1}{2} mv^2 = eV \] From this, we can find the velocity of the electron and proton. For the electron:
\[ v_e = \sqrt{\frac{2eV}{m_e}} \] For the proton:
\[ v_p = \sqrt{\frac{2eV}{m_p}} \] Now, the ratio of the de Broglie wavelengths of the electron and proton is:
\[ \frac{\lambda_e}{\lambda_p} = \frac{\frac{h}{p_e}}{\frac{h}{p_p}} = \frac{p_p}{p_e} = \frac{m_p v_p}{m_e v_e} \] Substituting the expressions for \( v_e \) and \( v_p \):
\[ \frac{\lambda_e}{\lambda_p} = \frac{m_p \sqrt{\frac{2eV}{m_p}}}{m_e \sqrt{\frac{2eV}{m_e}}} \] Simplifying:
\[ \frac{\lambda_e}{\lambda_p} = \sqrt{\frac{m_p}{m_e}} \] Thus, the correct answer is option (D) \( \sqrt{\frac{m_p}{m_e}} \).