An electron (mass m) and a proton (mass M) are accelerated with same potential difference, then

An electron (mass m_e ) and a proton (mass m_p ) are accelerated through the same potential, then lambda_e//lambda_p =

An electron of mass m_e and a proton of mass m_p are accelerated through the same potential difference. The ratio of the de Broglie wavelength associated with an electron to that associated with proton is

An electron of mass m_e and a proton of mass m_p are accelerated through the same potential difference. The ratio of the de Broglie wavelength associated with an electron to that associated with proton is

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

An electron and a proton are accelerated through the same potential difference. The ratio of their de broglic wavelengths will be

An electron and a proton are accelerated through the same potential difference. The ratio of their de-Broglie wavelengths will be

An electron and a proton are accelerated through the same potential difference. The ratio of their de broglic wavelengths will be

During an experiment an alpha -particle and a proton are accelerated by same potential difference , their de Broglie wavelength ratio will be : (Take mass of photon = mass of neutron)

An electron of mass m has de Broglie wavelength lambda when accelerated through a potential difference V . When a proton of mass M is accelerated through a potential difference 9V, the de Broglie wavelength associated with it will be (Assume that wavelength is determined at low voltage ) .