An electron in the beam of a typical television picture tube is accelerated through a potential difference of 2.0*10^4 V before it strikes the face of the tube. What is the energy of this electron, in electron volts, and what is its speed when it strikes the screen?
delta V = delta Ep/q
The Attempt at a Solution
well, I used the first equation to find its electric potential energy, which is (2.0*10^4)*(1.6*10^-19), and that gives me 3.2*10^-15J
then since it's asking in eV, and 1eV = 1.6*10^-19J, i do (3.2*10^-15)/(1.6*10^-19), which gives me 2*10^4eV (exact number as we started).
I don't see anything wrong with the procedure but the answer is 2*10^-4eV. what did i do wrong?
as for speed: (im using my 2*10^4eV as the energy)
Ek=0.5mv^2, and energy in J is 3.2*10^-15J
so 3.2*10^-15 = 0.5 (9.11*10^-31)v^2
that gives me 8.4*10^7m/s, which is exactly the same as the answer.
therefore i'm assuming the answer key for the energy in eV is wrong, what do you think?