Energy Needed to Remove an Electron

1. Jun 15, 2008

LHC

1. The problem statement, all variables and given/known data

If the total energy needed to remove an electron from a hydrogen atom is 24.6 eV, what is the distance that the electron is from the nucleus?

2. Relevant equations

I'm not really sure....I know that the answer is $$r = 5.85\times10^-11$$.

3. The attempt at a solution

As you can see, I know the answer, but I'm just not sure how to get to it. So far, I have thought of applying this process:

$$E_{e} = \frac{kq_{1}q_{2}}{r}$$

Technically, I get the answer if I sub in:$$q_{1}=q_{2}=e$$, $$E_{e}=24.6e$$, and solve for r.

Is this the correct process? I'm not really sure about the rationale behind it...lol I'm almost positive that there is another way.

Last edited: Jun 15, 2008
2. Jun 15, 2008

dynamicsolo

What you've written is the expression for the electrostatic potential energy between two charges q_1 and q_2 separated by a distance r. Yes, this is what you want to use.

Make sure you put everything into SI units, which means you'll need to convert that energy of 24.6 electron-volts into Joules...

3. Jun 16, 2008

LHC

Thanks for the help!

4. Jun 16, 2008

Andrew Mason

Your answer is correct. You can figure out the rationale from the definition of work. Work = Force x distance. Since the force changes with distance, you have to integrate (from beginning and end points, r to infinity):

$$W = \int_r^\infty F\cdot ds = \int_r^\infty -\frac{kQq}{r^2} dr = kQq/r - kQq/\infty$$

AM

5. Jun 16, 2008

Andrew Mason

Repeat of my above answer since Latex seems to be having problems:

$$W = \int_r^\infty F\cdot ds = \int_r^\infty \frac{kQq}{r^2} dr = \frac{kQq}{r} - \frac{kQq}{\infty}$$

AM