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#### Guest

##### Guest

I refer to the equation eV=1/2mv^2 where 'e' is the charge of the electron, V is the potential difference between two points, 'm' is the mass of the electron and 'v' is the speed of the electron.

Right the speed of light is taken as 3X10^8 so lets work with obtaining a speed of 4X10^8. Using 9.1X10^-31 as the mass of the electron when we substitute these values into the formula of kinectic energy 1/2mv^2 we get the value of 7.28X10^-14 which according to the above equation is equal to eV.

Now when we divide this value by 'e'(1.6X10^-19) we get 455000V as the potential difference. So to satisfy my curiousity what happens when you apply this potential difference to a cathode ray tube?