Question 10e38

Mar 29, 2016

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) ${F}_{c} = m \cdot {v}^{2} / r$

or equation 2)

2) ${F}_{c} = m \cdot {w}^{2} \cdot r$ (why? because $v = w \cdot r$. 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, $\frac{r e v}{s}$) .

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.

$1 N = 1 k g \cdot \frac{1 m}{s} ^ 2$

The angular velocity of the electron, $w$, is given in $\frac{r e v}{s}$
...but we want $\frac{r a d}{s}$!

So let’s convert 6×10^15(rev)/s to $\frac{r a d}{s}$. We simply multiply $r e v$ by $2 \pi$.

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) ${F}_{c} = m \cdot {w}^{2} \cdot r$
F_c=9.11×10^-31kg*(12pi×10^15(rad)/s)^2*5.3×10^-11m
F_c=6.86×10^-8N#