JEE MainPhysicsDual Nature of Matter
An electron is accelerated from rest through a potential difference V . A photon is found to have the exact same wavelength as the de Broglie wavelength of this accelerated electron. If m is the mass of the electron, e is its charge, and c is the speed of light, the ratio of the energy of the photon to the kinetic energy of the electron is:
Options
- AeV 2mc^2
- B2mc^2 eV
- C2mc^2 eV
- D1
Correct answer
C. 2mc^2 eV
Step-by-step solution
The kinetic energy of the electron accelerated through a potential difference V is: K = eV The de Broglie wavelength of this electron is: = h 2mK = h 2meV The energy of a photon having the same wavelength is: E_p = hc Substituting the expression for : E_p = hc ( h 2meV ) = c 2meV The ratio of the energy of the photon to the kinetic energy of the electron is: E_p K = c 2meV eV = 2mc^2(eV) (eV)^2 = 2mc^2 eV Answer: 2mc^2 eV