Saved Bookmarks
| 1. |
State de-Broglie relation for wavelength of matter waves. Show that the de-Broglie wavelength of electrons accelerated through a potential of V volt can be expressed as : lambda = h/sqrt(2meV) |
|
Answer» Solution :The de-Broglie wavelength associated with a material particle of mass m moving with velocity V is given by `lambda = h/(mv)` Wavelength associated with a moving electron. Suppose an electron STARTING from rest falls through a potential difference V Let charge on electron = e Mass of electron = m, Velocity acquired by the electron in FALLING through a potential difference, V = v Work done on the electron = EV Kinetic energy of the electron `=1/2mv^(2)` OBVIOUSLY, `1/2mv^(2) = eV` or `v = sqrt((2EV)/m)`........(1) It `lambda` is the wavelength associated with the electron, then according to de-Broglie wave equation. `lambda =h/(mv)` or `lambda =h/(msqrt((2eV)/m)) =h/sqrt(2meV)` |
|