The structure of NH3 the molecule is approximately that of an equilateral tetrah

Question

The structure of NH3 the molecule is approximately that of an equilateral tetrahedron, with three H+ ions forming the base and an N3- ion at the apex of the tetrahedron. The length of each side is 1.64 x10-10 m. Calculate the electric force that acts on each ion. (Let the H+ ions be in the x-y plane with H1 at (0, 0, 0), H2 at (a, 0, 0), and H3 at (a/2, av(3)/2, 0).The N3- ion, with charge q4 in our notation, is then at (a/2, a/2v(3), av(2/3)), where a = 1.64 x10-10 m. To simplify our calculations we'll set ke2/a2 = C = 8.56 x10-9 N. Use the following variable as necessary : C.)
*need to find the force on each electron (F1, F2, F3, and F4)*

Explanation / Answer

The structure of NH3 the molecule is approximately that of an equilateral tetrahedron, with three H+ ions forming the base and an N3? ion at the apex of the tetrahedron. The length of each side is 1.64 10-10 m. Calculate the magnitude of the electric force that acts on each ion. (Let the H+ ions be in the x-y plane with H1 at (0, 0, 0), H2 at (a, 0, 0), and H3 (a/2,(a*sqrt(3))/2,0), and the N is at (a/2,a/(sqrt(3)*2),a*sqrt(2/3))

where a = 1.64 10-10 m. To simplify our calculations we'll set (ke^2)/a^2 = C = 8.56 10^-9 N. Use the following variable as necessary : C.)

Find the magnitude of force on H1

2. Relevant equations
(K* lq1*q2l)/r^2=F(E)

3. The attempt at a solution

First of all after sketching a little diagram I concluded that the force of h2 on h1 was < - ,0,0>, h3 on h1 <-,-,0), and N on h1 <+,+,+>. Made these conclusions by looking at the direction the force would be. In the end I used these signs to give my magnitudes direction in order to add the vectors.

I found the distance between them all from h1 by using the distance formula. They have the same distance which is "a".

Then I set up triangles for them all except the first one because it is pushed straight down the x axis.

So for h2 on h1 I got (Ke^(2))/a^(2)<-1,0,0>

For h3 on h1 I got a triangle with a side in the x direction with a length of a/2 and the side in the y was (a*sqrt(3))/2. So I used tan(o/a)=30 degrees.

So I took the sin and cos of 30, and figured I needed to multiply my force by 1/2 to get the x comp, and root3 over 2 to get my y comp.
(Ke^(2))/a^(2)<-.5,-sqrt(3)/2,0>


Then for my N on h1 I did the same process and got

(3ke^2)/a^2<sqrt(3)/2,.5,.8164965809>

Then I added all my vectors together.

<(3^(3/2)-1-.5, 1.5-sqrt(3)/2, 3(.8164965809)>


The x and y component are apparently supposed to both be zero but I don't get them cancelling out.

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.