The de Broglie wavelength associated with an electron accelerated through a potential difference V is e and the de Broglie wavelength associated with a proton accelerated through the same potential difference is p. If their corresponding masses are me and mp, respectively, then the ratio of their de Broglie wavelengths ( e p ) is .
- mpme
- memp
- mpme
- ( mpme )2
Answer
(1) mpme
Solution
Kinetic energy gained: K=qV p= 2mK= 2mqV
= hp= h 2mqV
Electron and proton: same magnitude of charge e, same V
e p= mp eVme eV= mpme
Answer: (1)