# Electron orbital frequency of hydrogen atom if given orbit radius

• Mugen112
In summary: The force pointing toward the center (acceleration) would be found using Coulomb's force formula?F(or a in this case) = qE Divergence = qE/, so F=kq/r^2=v^2/r.
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.

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.

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.

Mugen112 said:
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

Mugen112 said:
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.

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

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 :)

## 1. What is the formula for calculating the electron orbital frequency of a hydrogen atom?

The formula for calculating the electron orbital frequency of a hydrogen atom is f = (1/2π)(k^2me^4/h^3), where k is the Coulomb's constant, me is the mass of the electron, e is the charge of the electron, and h is the Planck's constant.

## 2. How is the orbit radius of a hydrogen atom related to its electron orbital frequency?

The orbit radius of a hydrogen atom is directly proportional to its electron orbital frequency. This means that as the orbit radius increases, the electron orbital frequency also increases. This relationship is described by the formula f ∝ 1/r3, where f is the electron orbital frequency and r is the orbit radius.

## 3. Can the electron orbital frequency of a hydrogen atom be negative?

No, the electron orbital frequency of a hydrogen atom cannot be negative. It is a physical quantity that represents the number of revolutions per unit time of an electron around the nucleus. Since the number of revolutions cannot be negative, the electron orbital frequency also cannot be negative.

## 4. How does the electron orbital frequency of a hydrogen atom change with different orbit radii?

The electron orbital frequency of a hydrogen atom increases as the orbit radius decreases. This is because the electron is closer to the nucleus and experiences a stronger electrostatic force, resulting in a higher frequency of rotations. Conversely, as the orbit radius increases, the electron orbital frequency decreases.

## 5. What is the significance of knowing the electron orbital frequency of a hydrogen atom?

Knowing the electron orbital frequency of a hydrogen atom can provide valuable information about the atom's energy levels and the behavior of its electrons. It is also a fundamental aspect of atomic physics and is used in various calculations and experiments related to atomic structure and properties.

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