Question

Electrons are accelerated through a potential difference \( V_e \), and protons are accelerated through a potential difference of 4 V. The de-Broglie wavelengths are \( \lambda_e \) and \( \lambda_p \) for electrons and protons, respectively. The ratio of \( \frac{\lambda_e}{\lambda_p} \) is given by (given \( m_e \) is mass of electrons and \( m_p \) is mass of protons):

• A. \( \frac{\lambda_e}{\lambda_p} = \sqrt{\frac{m_e}{m_p}} \)
• B. \( \frac{\lambda_e}{\lambda_p} = \sqrt{\frac{m_p}{m_e}} \)
• C. \( \frac{\lambda_e}{\lambda_p} = \frac{1}{\sqrt{\frac{m_p}{m_e}}} \)
• D. \( \frac{\lambda_e}{\lambda_p} = 2 \sqrt{\frac{m_p}{m_e}} \)

Hint:

De-Broglie wavelength is inversely proportional to the square root of the mass of the particle. For electrons and protons, their wavelengths depend on their respective masses.

Answer 1

The Correct Option is D

The de-Broglie wavelength of a particle is given by the equation: \[ \lambda = \frac{h}{\sqrt{2m eV}} \] where \( h \) is Planck’s constant, \( m \) is the mass of the particle, \( e \) is the charge of the particle, and \( V \) is the potential difference.
For electrons and protons, their respective wavelengths are: \[ \lambda_e = \frac{h}{\sqrt{2m_e e V_e}}, \quad \lambda_p = \frac{h}{\sqrt{2m_p e V_p}} \] Since the potential difference is the same (4 V), the ratio \( \frac{\lambda_e}{\lambda_p} \) is: \[ \frac{\lambda_e}{\lambda_p} = \frac{\sqrt{m_p}}{\sqrt{m_e}} = 2 \sqrt{\frac{m_p}{m_e}} \]
Thus, the correct answer is (d).