Electric Potential Energy of electron orbits

[nckaytee]

An electron and a proton are initially very far apart (effectively an infinite distance apart). They are then brought together to form a hydrogen atom, in which the electron orbits the proton at an average distance of 5.43 * 10-11 m. What is EPEfinal - EPEinitial, which is the change in the electric potential energy?

I have no idea where to start

 

[LowlyPion]

An electron and a proton are initially very far apart (effectively an infinite distance apart). They are then brought together to form a hydrogen atom, in which the electron orbits the proton at an average distance of 5.43 * 10-11 m. What is EPEfinal - EPEinitial, which is the change in the electric potential energy?

I have no idea where to start

Isn't the potential energy given by PE = kq1*q2/r

If EPE at infinity is 0, then ... The change is ...

 

[nckaytee]

In my notes I have $$\Delta E.P.E$$ of electron = q(Vb-Va) = (-e)(Vb-Va)

I am really not getting this


Im sorry, I might have something here... One minute..

 

[nckaytee]

Okay I am confused

 

[LowlyPion]

In my notes I have $$\Delta E.P.E$$ of electron = q(Vb-Va) = (-e)(Vb-Va)

I am really not getting thisIm sorry, I might have something here... One minute..

What is the charge of an electron and a proton?

e = –1.602176487(40) × 10–19C
p = 1.60217653(14)×10−19 C

 

[nckaytee]

So, would it be...

(-1.6*10^19)(0 - 5.43 *10^-11)

 

[LowlyPion]

So, would it be...

(-1.6*10^19)(0 - 5.43 *10^-11)

It's the product of both charges.

PE = kq1*q2/rAnd check the exponent of the charges. And don't forget k.

 

[nckaytee]

Okay, so I did PE = [(9*10^9)(-1.6*10^-19)(1.6*10^-19)] / 5.43*10^-11

My answer is right! -4.24 * 10^18

Thank you

 

[LowlyPion]

Okay, so I did PE = [(9*10^9)(-1.6*10^-19)(1.6*10^-19)] / 5.43*10^-11

My answer is right! -4.24 * 10^18

Thank you

You're welcome.

And don't act so surprised. You had it nailed from kqq/r.

Good luck.