A proton and an alpha particle are accelerated from rest through the same potential difference V . The ratio of their de-Broglie wavelengths ( _p : _ ) is

Step-by-step solution

The kinetic energy acquired by a particle of charge q when accelerated from rest through a potential difference V is K = qV . The de-Broglie wavelength is given by = h 2mK = h 2mqV . For a proton: mass m_p = m , charge q_p = e . For an alpha particle: mass m_ = 4m , charge q_ = 2e . The ratio of their de-Broglie wavelengths is: _p _ = m_ q_ m_p q_p Substituting the values: _p _ = (4m)(2e) (m)(e) = 8 = 2 2 Thus, the ratio is 2 2 :1 . Answer: 2 2 :1