Question #10e38

Question

Question #10e38

1 Answer

Impact of this question

2411 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-03-29T21:11:59

The centripetal force acting on the electron, allowing it to maintain its circular orbit around the nucleus, is $F_{c} = 6.86 \times 10^{- 8} N$ $F_c=6.86×10^-8N$ .

Explanation:

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

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) $F_{c} = m \cdot \frac{v^{2}}{r}$ $F_c = m*v^2/r$

or equation 2)

2) $F_{c} = m \cdot w^{2} \cdot r$ $F_c = m*w^2*r$ (why? because $v = w \cdot r$ $v=w*r$ . Verify this for yourself!)

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

In equation 2, $w$ $w$ 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, $\frac{r e v}{s}$ $(rev)/s$ ) .

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

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

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

$1 N = 1 k g \cdot \frac{1 m}{s^{2}}$ $1 N = 1kg*(1m)/s^2$

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

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

$w = \frac{2 π \cdot r a d}{1 \cdot r e v} \cdot 6 \times 10^{15} \frac{r e v}{s} = 12 π \times 10^{15} \frac{r a d}{s}$ $w=(2pi*rad)/(1*rev)*6×10^15(rev)/s=12pi×10^15(rad)/s$

So now, that our units are consistent, substituting

$m = 9.11 \times 10^{- 31} k g$ $m=9.11×10^-31kg$ ,
$w = 12 π \times 10^{15} \frac{r a d}{s}$ $w=12pi×10^15(rad)/s$ ,
and $r = 5.3 \times 10^{- 11} m$ $r=5.3×10^-11m$ into equation 2 we get.

3) $F_{c} = m \cdot w^{2} \cdot r$ $F_c=m*w^2*r$
$F_{c} = 9.11 \times 10^{- 31} k g \cdot {(12 π \times 10^{15} \frac{r a d}{s})}^{2} \cdot 5.3 \times 10^{- 11} m$ $F_c=9.11×10^-31kg*(12pi×10^15(rad)/s)^2*5.3×10^-11m$
$F_{c} = 6.86 \times 10^{- 8} N$ $F_c=6.86×10^-8N$

Science

Math

Humanities

... and beyond

Question #10e38

1 Answer

Explanation:

Related questions

Impact of this question