Electron orbital frequency of hydrogen atom if given orbit radius

[Mugen112]

Homework Statement

In a classical model of the hydrogen atom, the electron moves around the proton in a circular orbit of radius 0.053 nm.

A) What is the electron's orbital frequency?

Homework Equations

F = qE
E= kq/r^2
angular velocity = v^2/r

The Attempt at a Solution

I'm developing a sever hate for Physics. This problem seems to be an easier problem, yet I still can't seem to get it. This is the only class that offers no support in terms of answering questions. I've read the chapters several times, yet I still have trouble seeing how the chapter and the questions after the chapter correlate. Again, this problem is an easier problem of the bunch. I don't see how the book not offering any help would be beneficial at ALL to learning the material. I learned previous chapters (kinematics/gravity/friction) all with great help from the chapter AND I understood how they formulated all their equations. I remember them and I know how to use them in a diverse aray of situations. I truly believe that they took the same questions from the previous book (Physics for Scientists and Engineers by Knight 1st ed.) and took a lot of the context out of the chapters. I know that those of you that have learned the material will probably say that it is the only way to learn this material... but I find that incredibly hard to believe. Am I the only one that thinks this or is this Physics book just put together poorly? I could write a book about how much I really hate this class... BUT ANYWAY... sorry for the vent...

So let's see here... they give the distance between the proton and electron. I have NO idea why in the hell an electron would orbit a proton (I know it does... but OK). From the fact that it is a hydrogen atom implies that there is only one proton and one electron of charge -e and e (1.16 x 10^-19 C) .

Now, because the electron orbits the proton, I suppose we would use the angular velocity formulas. The force pointing toward the center (acceleration) would be found using Coulomb's force formula?

a = v^2/r

F(or a in this case) = qE
E= Kq/r^2

so.. Kq^2/r^2 = v^2/r ?

I plug everything in and I get a number for tangential velocity to equal 1.5 x 10^-9 m/s. Then the circumference of the orbit is 2pi()(radius). I get that number... Then Plug both of those into D=RT.. solve for T, find out how many revolutions persecond for the frequency? Mastering Physics says its wrong. I give up.

 

[dark adonis]

It looks like you are very close:
You made an error by setting the Coulomb Force equal to the acceleration, you forgot that it should be the centripetal force which should be
$$K\frac{q^2}{r^2}=\frac{m v^2}{r}$$ which you solve as you did.

 

[Mugen112]

Ok, so.. I tried it again and it's still wrong.
For V, I get 1583733 m/s with the equation that you gave me.

V= sqrt (((9*10^9)((1.16*10^-19)^2)) / ((.053 * 10^-9)(9.11*10^-31)) = 1583733 m/s

Then 2pi()R = 3.33*10^-10

D = RT

3.33*10^-10 / 1583733 = 2.1 * 10^-16 s

so that's 1 revolution in 2.1 * 10^-16? So I'm guessing taking the inverse of this will give you the number of revolutions per second = 4.8 *10^15 Hz. MasteringPhysics still says this is wrong. Forgot to say thanks for the help tho.

 

[dark adonis]

Ok, so.. I tried it again and it's still wrong.
For V, I get 1583733 m/s with the equation that you gave me.

V= sqrt (((9*10^9)((1.16*10^-19)^2)) / ((.053 * 10^-9)(9.11*10^-31)) = 1583733 m/s

Then 2pi()R = 3.33*10^-10

D = RT

3.33*10^-10 / 1583733 = 2.1 * 10^-16 s

so that's 1 revolution in 2.1 * 10^-16? So I'm guessing taking the inverse of this will give you the number of revolutions per second = 4.8 *10^15 Hz. MasteringPhysics still says this is wrong. Forgot to say thanks for the help tho.

Well your analytic expression for v looks correct : $$v=\sqrt{\frac{K e^2}{r m_e}}$$ and your relationship between v and period is right $$T = 2 \pi r/v$$
so the only thing possibly wrong is numerical precision, have you tried entering more significant digits? I know with the system in use at my school the computer's evaluation is based purely on percents rather than significant digits... I'm terribly sorry that I can't help any further, but everything looks right.

Unless I've forgotten something this looks right, the only problem I can see is that you have an error of

 

[mikehollowed]

From the fact that it is a hydrogen atom implies that there is only one proton and one electron of charge -e and e (1.16 x 10^-19 C)

I know it doesn't matter now, because this was like 2 years ago, but e is -1.6 x 10^-19.

 

[Abelard]

Bohr radius is 5.29e-11m by the way.

 

[Lorenzo532]

You are on the right track but,
Fe= Fc
Ke^2/r^2 = mv^2/r
r=Ke^2/mv^2
V^2=Ke^2/mr
∴V^2=√((9x10^9)x((1.6x10^-19)^2)/(9.11x10^-31)x(5.29x10^-11)
V= 2'186'340.09 ms-1
∴V= 2.2x10^6ms-1

∴Frequency =
2∏r/v
t= (2x∏x (5.29x10^-11))/(2.2x10^6)
∴t= 1.51x10^-16 seconds

∴f= 1/(1.51x10^-16)
∴f= 6.6x10^15 revolutions per second

There you go :)