91Ó°ÊÓ

Distance of center from each hydrogen atom, $\mathrm{d}=9.40×{10}^{-11}\mathrm{m}$

Distance of hydrogen atom from nitrogen atom,${\text{³¢=10.14×10}}^{-11}\mathrm{m}$

For a system of particles, the whole mass of the system is concentrated at the center of mass of the system.

By the symmetry ofthemolecule, it is clear that the x coordinate of the center of mass will be ontheplane containing hydrogen molecules.Using the given structure of H₃ molecule, we can find the distance between nitrogen molecules totheplane containing all hydrogen atoms. From that distance, we can find the y coordinate of center of mass of nitrogen molecule.

The expression for the coordinates of the center of mass are given as:

${\stackrel{→}{r}}_{com}=\frac{1}{M}\underset{i=1}{\overset{n}{∑}}{m}_i{r}_i$ … (i)

Here, $M$is the total mass, $m_i$ is the individual mass of ith particle and $r_i$ is the coordinates of ithparticle.

From the symmetry of the molecule, we can directly conclude that the center of mass will be on the plane containing the hydrogen atoms. Therefore, the x coordinate of the center of mass will be at the origin.

$x_{com}=0$

Thus, the x coordinate of the center of mass is 0.

First we found the distance from nitrogen to plane containing hydrogen atoms.

We can use Pythagoras theorem here.

${\mathrm{y}}^2\mathrm{n}+{\mathrm{d}}_2={\left(\mathrm{L}\right)}^2$

Substitute the values in the above equation.

$$\begin{gathered} {\mathrm{y}}_{\mathrm{N}}=\sqrt{{(10.14×{10}^{-11}\mathrm{m})}^2-{(9.40×10-11\mathrm{m})}^2} \\ =3.803×{10}^{-11}\mathrm{m} \end{gathered}$$

Now, the y coordinate of the center of mass can be calculated as:

$$\begin{gathered} {\mathrm{y}}_{\mathrm{com}}=\frac{{\mathrm{m}}_{\mathrm{N}}{\mathrm{y}}_{\mathrm{N}}}{{\mathrm{m}}_{\mathrm{H}}+3{\mathrm{m}}_{\mathrm{H}}} \\ =\frac{(14.0067\mathrm{u})(3.803×{10}^{-11})}{(14.0067\mathrm{u})+3(1.0067\mathrm{u})} \\ =3.13×{10}^{-11}\mathrm{m} \end{gathered}$$

Thus, the $y$coordinate of the center of mass is $3.13×{10}^{-11}m$.