Answer to Question #142002 in Quantum Mechanics for Joe

Question

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;

Answer 1

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 $a$ is

$$x_c=\frac{\sqrt{3}}{6}a,$$

The Y-coordinate for our disposition is

$$y_c=0.$$

What left is to find $a$ and $z_c$.

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

$$a=2\cdot0.10\cdot\text{sin}54°=0.16\text{ nm}.\\\space\\ x_c=\frac{\sqrt{3}}{6}a=\frac{\sqrt{3}}{6}0.16=0.046\text{ nm}.$$

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:

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

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

$$z_c=\frac{z_{\text{H}_3}\cdot m_{\text{H}_3}+z_{\text{N}}\cdot m_{\text{N}}}{m_{\text{H}_3}+m_{\text{N}}}=\frac{0+0.038\cdot14}{1\cdot3+14}=0.032\text{ nm}.$$

Therefore, the center of gravity has the following coordinates:

$$x_c=0.046\text{ nm},\\ y_c=0,\\ z_c=0.032\text{ nm}.$$