Electrons are accelerated in a television tube through a potential difference of 8.5 kV. What is the highest frequency of the electromagnetic waves emitted when these electrons strike the screen of the tube?
λ = c/f
E (of the electron) = V*e
E (of the photon) = h*c/λ
The Attempt at a Solution
Without knowing the work function of the tube, the best I could come up with was to assume that all the kinetic energy of the electron was transferred to the photon and set the energies equal to one another. Simple algebra substituting the wavelength for it's definition in terms of frequency gives the equation:
f = V*e/h
Substitution of numeric values then gives
EDIT: I accidentally used h-bar instead of h. It still comes out different from the desired answer though.
f = 8500*1.6E-19/4.14E-15 = .329 Hz
The online homework I have says that I am off by 18 orders of magnitude: the correct answer is 2.1E18
Where am I going wrong here?