Question

What is the wavelength of an electron accelerated through a 28.5 kV potential, as in a TV tube?

Answer 1

First we need to equate qV (accelerating potential) with the K.E. the particle will have (1/2 m v^2)-:
Kinteic Energy (of Electron) = q V = 1/2 m v^2
(This is neglecting reletivistic effects)

K.E. = q V = 1.6 x 10^-19 x 28500

K.E. = 3.04 x 10^-15 = 1/2 m v^2

Now we need velocity 'v' (so we re-arrange this for v^2)-:

v^2 = 3.04^-15 x 2 / m

v^2 = 3.04^-15 x 2 / 9.11 x 10^-31

V^2 = 6.674 x 10^15
v = 8.169 x 10^7 ms^-1 (this is ~27% of the speed of light)

Now we can work out the De Broglie wavelength (just put numbers in)-:

lambda = 6.63 x 10^-34 / (9.11 x 10^-31) x (8.169 x 10^7)

lambda = 8.909 x 10^-12 metres