# Ionization Energy Help

#### Bt98

Hello everyone, this is my first time posting here so apologies if this is not the appropriate place to put this. I was going through exam review and found a question that is tripping me up. It is probably a pretty simple solution and I just can't see it because I've been looking at it too long. Here it is:

In the Bohr model of the hydrogen atom, what is the smallest amount of work that must be done on the electron to move it from its circular orbit , with a radius of 0.529x10^-10m, to an infinite distance from the proton? This value is referred to as the ionization energy of hydrogen.

Here is where I am:

I used the relationship between centripetal and electrostatic force to find the velocity of the electron (2,185,309ish m/s).

Work=(Kf+Uf)-(Ki+Ui), but because the electron is being moved to an infinite distance away both final energy values=0, so W=0-(Ki+Ui).

I treat Ui as =kq²/r, since U=Vq and V=kq/r. K is straightforward enough

Combining all this, I end up at W=0-((mv²/2)+(kq²/r)). However, the solution to the problem shows that kq²/r should be negative in this case, ie W=0-((mv²/2)+(-kq²/r)). The latter solution is correct, as it comes out to 2.18x10^-18J.

So my question is, why is kq²/r negative there? Again it is probably really simple, but for the life of me I can't seem to make sense of it. Any explanation would be appreciated. Thanks!

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#### Woody

I may be treating this too simple mindedly but...

½mv² is a measure of the kinetic energy of the electron
which will be acting to move the electron out of its orbit.

kq²/r is a measure of the electrostatic energy
which will be acting to move the electron toward the centre of the orbit.

Thus they are acting in opposite directions
thus they have opposite signs.