Calculate the wavelength and kinetic energy of a beam of ele

Calculate the wavelength and kinetic energy of a beam of electrons accelerated from rest by a voltage increment of 100 V and the kinetic energy of an electron with a wavelength of 2 A. Using the uncertainty principle, estimate the ground state energy of an electron confined in a one-dimensional wire of length 1 nm.

Solution

The kinetic energy of an electron accelerated through a potential difference of V volts is given by the equation:

½ mv² = eV where e is the electron charge (1.6x10^-19 C) [You must be given the electron charge and Planck\'s constant in order to answer this question].

and so v = (2eV/m) = [1.6*10^-19 *100 ]^0.5 / [9.1*10^-31 kg]^0.5 = 4.19*10^6 m/s

Therefore: wavelength = h/mv = 6.625*10^-34/[9.11x10^{- 31}x4.19x10⁶] = 1.737x10^-10 = 0.1737 nm

wavelength = h/ mv

Given wavelength = 2 A = 2*10^-10 m

v =  h / m*wavelength = 3.64*10^6 m/s

K.E = 0.5*mv^2 = 6.02893*10^-18 J

C]

dx. dp = h/4pi

dp *10^-9 = h/4pi

dp = 5.2*10^-25

E = p^2/2m = [5.2*10^-25]^2 / 2*9.1*10^-31 = 1.485*10^-19 J