Question #10e38

1 Answer
Mar 29, 2016

Answer:

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#