Question

Question #10e38

Answer 1

The centripetal force acting on the electron, allowing it to maintain its circular orbit around the nucleus, is `F_c=6.86×10^-8N`.

Explanation:

Here, we are asked to calculate the ’sum’ of the centripetal forces on an electron in a circular orbit of ‘presumed’ radius `r=5.3 × 10^-11m`.

For this problem, we can assume a simple model for the hydrogen atom that involves one particle (the electron) traveling in a circular orbit around another more massive particle (the hydrogen nucleus)...much like the moon orbits the earth!

If this is the case, the centripetal force can then be described by equation 1

1) `F_c = m*v^2/r`

or equation 2)

2) `F_c = m*w^2*r` (why? because `v=w*r`. Verify this for yourself!)

here `v` represents the tangential velocity of an object (having a mass `m`) traveling around a circle of radius `r`.

In equation 2, `w` is the angular velocity of the object. These formulas for centripetal force, are equivalent, but using the second form makes the solution a bit more straight forward (since the angular velocity of the electron is given, `(rev)/s`) .

At this point, finding the amount the centripetal force simply requires substituting the numerical values for the physical quantities `m`, `w` and `r` into equation 2.

But before we do that, let’s make sure that the units of the physical quantities `m`, `w` and `r` are consistent with N. Currently, `w` is not. So let’s fix this!

Ultimately, in the end, we want a force in newtons, N.

`1 N = 1kg*(1m)/s^2`

The angular velocity of the electron, `w`, is given in `(rev)/s`
...but we want `(rad)/s`!

So let’s convert `6×10^15(rev)/s` to `(rad)/s`. We simply multiply `rev` by `2pi`.

`w=(2pi*rad)/(1*rev)*6×10^15(rev)/s=12pi×10^15(rad)/s`

So now, that our units are consistent, substituting

`m=9.11×10^-31kg`,
`w=12pi×10^15(rad)/s`,
and `r=5.3×10^-11m` into equation 2 we get.

3) `F_c=m*w^2*r`
`F_c=9.11×10^-31kg*(12pi×10^15(rad)/s)^2*5.3×10^-11m`
`F_c=6.86×10^-8N`