Determining radius of electron's orbit Need help .

[gookies]

Homework Statement

A positively charged Helium atom has two protons in the nucleus and one electron in the shell, in a classic model of the atom, this electron is in a circular orbit around the nucleus with an angular momentum of 4.718x10^-34 Js. What is the radius of the orbit?

Homework Equations

L = sqrt k*e²*m*r_o

The Attempt at a Solution

Using that equation, I've tried solving for the radius, but I'm not getting the correct answer. The professor told us it had to do with this equation, so I'm not sure what I'm doing wrong. I've gotten 2.37*10^-10m as my answer.

 

[collinsmark]

Hello gookies,

Welcome to Physics Forums!

L = sqrt k*e²*m*r_o

Using the specified classic model, the above equation might apply to a neutral hydrogen atom, but not a positively charged helium atom.

Try to re-derive the equation for a positively charged helium atom. Note that,

F_e = k\frac{q_1 q_2}{r_0^2}

F_{centrepital} = \frac{mv^2}{r_0}

And if the velocity is perpendicular to the radius,

\vec L = \vec r_0 \times m \vec v \ \rightarrow \ L = r_0mv

Then substitute in the charges accordingly (remember the helium atom has two protons, not one).

 

[gookies]

Ah. I'm still confused. So do you have to solve for the velocity first and substitute F for mv^2/r? So the end equation would be, mv^2/r = k q1q2/r^2? And with the charges, would q1 be the charge of the electron and q2 would be the charge of 2 protons?

 

[collinsmark]

Ah. I'm still confused. So do you have to solve for the velocity first and substitute F for mv^2/r? So the end equation would be, mv^2/r = k q1q2/r^2?

That's what I did anyway, yes.

After you solve for the velocity, modify it accordingly to make it into an expression for angular momentum.

And with the charges, would q1 be the charge of the electron and q2 would be the charge of 2 protons?

Yes, that's right. Or you could make q₁ be the charge of the two protons (in the nucleus) and q₂ be the charge of the single electron. Just pick one to be the nucleus and the other to be the electron. Which one is which doesn't matter.

 

[gookies]

Okay, I just looked at it and you don't have to solve for velocity. I changed L=r*m*v to solve for v and substituted v in the equation of F = mv^2/r. I set that equal to kqq/r^2 and then I eventually set everything equal to L first, then solved for r. my final equation was, L^2 = k*q*q*r*m. Would that be correct, Collin?

Edit: Realized it's exactly like the equation I gave. :( Not sure what to do.

 

[collinsmark]

Okay, I just looked at it and you don't have to solve for velocity. I changed L=r*m*v to solve for v and substituted v in the equation of F = mv^2/r. I set that equal to kqq/r^2 and then I eventually set everything equal to L first, then solved for r. my final equation was, L^2 = k*q*q*r*m. Would that be correct, Collin?

Yes, your way will work too.

Edit: Realized it's exactly like the equation I gave. :( Not sure what to do.

It's almost the same. In the equation you derived there is a q₁q₂ instead of your original equation that had an e². The important point here is that in a positive helium ion, q₁ ≠ q₂ (remember, the nucleus of the helium ion has two protons, not one).

Plug in the charges and see what you get.

 

[gookies]

Ah, it works now. I guess it was just plugging in my numbers that screwed me. Thanks Collin!