Answer to Question #142002 in Quantum Mechanics for Joe

Question #142002
A simple model of ammonia consists of a pyramid of hydrogen atoms at the vertices of the equilateral base and the nitrogen atom at the apex. The N-H distance is 0•10 nm and the angle between two N--H bonds is 108
Find the center of gravity of the molecule. (Atomic weights: H N 14;
1
Expert's answer
2020-11-18T14:51:57-0500

Put this molecule in a 3d coordinate system so that the midpoint between two hydrogen atoms is at 0, and the line that crosses these two hydrogens is lying on a Y-axis so that the third hydrogen is on X-axis. Nitrogen atom is in positive Z-direction. We know that the centre of gravity of an equilateral triangle of side aa is


xc=36a,x_c=\frac{\sqrt{3}}{6}a,

The Y-coordinate for our disposition is


yc=0.y_c=0.

What left is to find aa and zcz_c.

But on the edge of any triangle HNH we have a right triangle with base a/2a/2, hypothenuse of 0.10 nm, and angle of 108°/2=54°.108°/2=54°. So, we can find aa:


a=20.10sin54°=0.16 nm. xc=36a=360.16=0.046 nm.a=2\cdot0.10\cdot\text{sin}54°=0.16\text{ nm}.\\\space\\ x_c=\frac{\sqrt{3}}{6}a=\frac{\sqrt{3}}{6}0.16=0.046\text{ nm}.

Regarding the z-coordinate: consider this molecule as a dumbbell with 3 hydrogens on one side and one nitrogen on the other, the distance between the 'weights' is the height of the pyramid:


h=0.12(33a)2=0.038 nm.h=\sqrt{0.1^2-\bigg(\frac{\sqrt3}{3}a\bigg)^2}=0.038\text{ nm}.

Thus, the z-coordinate of the centre of gravity of the molecule is

zc=zH3mH3+zNmNmH3+mN=0+0.0381413+14=0.032 nm.z_c=\frac{z_{\text{H}_3}\cdot m_{\text{H}_3}+z_{\text{N}}\cdot m_{\text{N}}}{m_{\text{H}_3}+m_{\text{N}}}=\frac{0+0.038\cdot14}{1\cdot3+14}=0.032\text{ nm}.

Therefore, the center of gravity has the following coordinates:

xc=0.046 nm,yc=0,zc=0.032 nm.x_c=0.046\text{ nm},\\ y_c=0,\\ z_c=0.032\text{ nm}.


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!

Leave a comment