Question

An electron of mass \(m_e\) and a proton of mass \(m_p\) are accelerated through the same potential. Then the ratio of their de Broglie wavelengths is

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

Hint:

When particles with the same charge are accelerated through the same potential, the lighter particle (electron) will always have a longer de Broglie wavelength than the heavier one (proton).

Answer 1

Answer: (E)

Concept: The de Broglie hypothesis associates a wave nature with every moving particle.
• Wavelength Equation: The de Broglie wavelength (\(\lambda\)) is given by \(\lambda = \frac{h}{p}\), where \(h\) is Planck's constant and \(p\) is momentum.
• Momentum and Kinetic Energy: Momentum is related to kinetic energy (\(K\)) by the relation \(p = \sqrt{2mK}\).
• Acceleration through Potential: When a charge \(q\) is accelerated through a potential \(V\), its kinetic energy is \(K = qV\).

Step 1: Derive the general expression for wavelength in terms of potential.
Substituting \(K = qV\) into the momentum-wavelength relation: \[ \lambda = \frac{h}{\sqrt{2mqV}} \] Since both the electron and proton have the same magnitude of charge (\(e\)) and are accelerated through the same potential (\(V\)), the wavelength is inversely proportional to the square root of the mass: \[ \lambda \propto \frac{1}{\sqrt{m}} \]

Step 2: Calculate the ratio of wavelengths.
Let \(\lambda_e\) be the wavelength of the electron and \(\lambda_p\) be the wavelength of the proton: \[ \frac{\lambda_e}{\lambda_p} = \frac{\frac{h}{\sqrt{2m_eeV}}}{\frac{h}{\sqrt{2m_peV}}} \] Canceling common terms (\(h, 2, e, V\)): \[ \frac{\lambda_e}{\lambda_p} = \frac{\sqrt{m_p}}{\sqrt{m_e}} = \sqrt{\frac{m_p}{m_e}} \]