Relating final speed of electron to its charge mass and voltage difference

[franbella]

Assuming that the speed of the electrons is zero as they are emitted from the filament and that as they reach the anode it is v, find an expression relating the final speed v to the charge on an electron, the mass of an electron and deltaV, and hence calculate the speed v of the electrons after passing through a voltage difference of 120 kV.


I'm new to this, but the equation I have come up with is
v = √ 2QΔV
m

I then get this:

√ 2 x (1.602 x 10-19J) x (120-0 x 103eV) = √ 3.8448 x 10-14
8.2 x 10-14 J 8.2 x 10-14 J
√0.46887804 = 0.6847466977
To two significant figures and in scientific notation this is 6.8 x 10-1 eV s-1

I think I'm tying myself in knots - please help!

 

[Thaakisfox]

You should divide by the mass under the square root, and do care about the units.

 

[franbella]

Sorry, my first post doesn't make it look like this because my m is not under the square root, but it should be - just an error with copy and paste. So I did do this, but I still feel lost. Is the equation I'm using applicable - the number's I've come up with don't seem right.

 

[Thaakisfox]

i noticed that in your original formula you put the m. What i meant was that when you put in the numbers i don't see you dividing by m.

 

[Orangeman]

of the electrons after passing through a voltage difference of 120 kV.


I'm new to this, but the equation I have come up with is
v = √ 2QΔV
m

I then get this:

√ 2 x (1.602 x 10-19J) x (120-0 x 103eV)

Little hard to see from post but remember to change kV into V and charge is in C. Therefore C x V = J, base units of J= kg m2 s-2.

kg m2 s-2
kg

= m2 s-2

square root of m2 s-2 = m s-1

Hope this helped.

 

[franbella]

Thank you - good reminder - I always forget. Think am getting somewhere now.