We need to find the work done by the proton and the other two electrons on one electron in moving it in from infinity to its final location.. Then triple it.
I will call a side of the triangle s
You can put in the nanometer later.
I will call proton +q, electron -q
Now in final configuration how far is our electron from the proton?
xp = 2/3 altitude = (2/3) (s/2)sqrt 3)
I am just going to call it xp
now the work done on the electron moving from xp to pull it from there to infinity is its potential energy at infinity. In other words the PE is negative at xp if 0 at infinity for our problem with the proton (not so when we get to the electrons)
force = k (+q)(-q)/x^2
Pe =integral k q^2 dx/x^2 from xp to oo
= - k q^2 /xp
Now for the electrons (we have to push them together, compress the spring, so this PE will be +)
for each, final distance = s
so for each
PE = +k q^2/s and we have two of them
so
final PE = 2 k q^2/s - k q^2/xp
but xp = (2/3) s
so
PE = k q^2 ( 2/s - 1.5/s)
= (1/2)k q^2/s
CHECK MY ARITHMETIC!!!!
Three electrons form an equilateral triangle 1.00 nm on each side. A proton is at the center of the triangle.What is the potential energy of this group of charges?
3 answers
Oh, do not forget to triple it, that was all just for pulling the first electron in.
Oh dear, xp is not 2/3 s but (2/3) (s/2)sqrt 3
which is
(s/3) sqrt 3
so
final PE = k q^2 (2/s - 1/[(s/3) sqrt 3]
= k q^2 ( 2/s - 3/(s sqrt 3))
=(k q^2/s) ( 2-1.73 )
=.27 k q^2/s
multiply by 3 to get total
=.81 k q^2/s
I have a + potential energy here which makes me think I either have an error or you little atom here will fall apart.
Normally the potential energy you lose whne an electron approaches a proton results in kinetic energy of its orbit around the proton (Bohr model of hydrogen atom).
OH - Of course it is unstable. We need three protons to hold three electrons in stable orbit!
Oh dear, xp is not 2/3 s but (2/3) (s/2)sqrt 3
which is
(s/3) sqrt 3
so
final PE = k q^2 (2/s - 1/[(s/3) sqrt 3]
= k q^2 ( 2/s - 3/(s sqrt 3))
=(k q^2/s) ( 2-1.73 )
=.27 k q^2/s
multiply by 3 to get total
=.81 k q^2/s
I have a + potential energy here which makes me think I either have an error or you little atom here will fall apart.
Normally the potential energy you lose whne an electron approaches a proton results in kinetic energy of its orbit around the proton (Bohr model of hydrogen atom).
OH - Of course it is unstable. We need three protons to hold three electrons in stable orbit!
I wish they had given you three protons at the center.
Then we could have figured out how much PE was lost by each electron in being pulled in toward the protons from infinity. We could have then converted that to (1/2) m v^2 and figured out how fast the electrons were going when they got to within 1 nanometer.
Then we could have figured out how much PE was lost by each electron in being pulled in toward the protons from infinity. We could have then converted that to (1/2) m v^2 and figured out how fast the electrons were going when they got to within 1 nanometer.
1 answer