# I Electron in orbit of around a single proton

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1. Jan 13, 2017

### Ian Baughman

Today I was doing some reading and I came across this topic. If we have a stationary hydrogen atom with a single electron in orbit around the nucleus and want to calculate the kinetic energy of the electron we would take the following approach.

1) Using Newton's second law:
F = ma ⇒ FE = mac ⇒ k(q1q2)/r2 = mv2/r​
2) We know:
charge of the electron = -e = -1.602×10-19 and the charge of the proton = +e = 1.602×10-19
3) Now using equation in step 1:
-k(e2)/r = mv2 ⇒ -k(e2)/(2r)=(1/2)mv2

We know kinetic energy is suppose to be positive so why is it that it comes out negative in this case?

2. Jan 13, 2017

### BvU

3. Jan 13, 2017

### Ian Baughman

Towards the nucleus of the atom.

4. Jan 13, 2017

### BvU

So FE provides the centripetal force. It's the only force around. $mv^2\over r$ also points towards the center, so it needs a minus sign too.