An electron (mass m_e ) and a proton (mass m_p ) are initially at rest. Both particles are then accelerated through the exact same potential difference V . What is the ratio of the de Broglie wavelength of the electron to that of the proton ( _e / _p ) ?

Step-by-step solution

The de Broglie wavelength of a particle is given by = h p The kinetic energy K acquired by a particle of charge q accelerated through a potential difference V is K = qV The momentum p is related to kinetic energy by p = 2mK = 2mqV For an electron and a proton, the magnitude of charge is the same, q = e The de Broglie wavelengths are _e = h 2m_e eV and _p = h 2m_p eV Taking the ratio gives _e _p = m_p m_e Answer: m_p m_e