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?
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!!!!
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!
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.
To calculate the potential energy of this group of charges, we need to consider the interaction between the charges. The potential energy is given by the formula:
U = ke * (q1 * q2) / r
where U is the potential energy, ke is the electrostatic constant (8.99 × 10^9 N m^2/C^2), q1 and q2 are the charges, and r is the distance between the charges.
In this case, we have three electrons (each with a charge of -1.6 × 10^-19 C) forming an equilateral triangle with a side length of 1.00 nm (1.00 × 10^-9 m). The proton at the center has a charge of +1.6 × 10^-19 C.
Now, let's consider the potential energy between each electron and the proton. Since the triangle is equilateral, the distance between each electron and the proton is the same (r = 1.00 nm).
Calculating the potential energy for one pair of charges:
U1 = ke * (q1 * q2) / r
= (8.99 × 10^9 N m^2/C^2) * (-1.6 × 10^-19 C * 1.6 × 10^-19 C) / (1.00 × 10^-9 m)
Since there are three pairs of charges (electron-proton interactions), we need to multiply this result by 3 to get the total potential energy:
U_total = 3 * U1
By plugging in the values and calculating, we can find the potential energy of this group of charges.
Note: It's important to consider that the potential energy of this system may change if the electrons are in motion or if there are other external forces acting on the charges.