Electron in orbit of around a single proton

[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 ⇒ F_E = ma_c ⇒ 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?

 

[BvU]

Which way is F_E pointing ?

 

[Ian Baughman]

Towards the nucleus of the atom.

 

[BvU]

So F_E 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.