Fine the de-Broglie wave length related to an electron accelerated by 104 volt.
Answers
Answer:
The de Broglie expression for wavelength of a moving particle in quantum mechanics is wavelength = (Planck’s constant/momentum of the particle), that is, lambda = h/p. The momentum p can be obtained by equating the kinetic energy of the particle to the acceleration potential V. The kinetic energy of the particle is 1/2 mv2 (half *mass of the particle * square of the velocity). If the charge of the electron is e, then, the accelerating potential V gives the electron an energy equivalent to eV. By equating the kinetic energy to this, we get, lambda as, (h/square root (2meV)). Substituting the appropriate values for Planck’s constant, electron charge and mass, the wavelength would be approximately 1.02 Angstrom. Hope this was not a homework problem.