Calculating Interatomic Distance from Spherical Coordinates

The problem requires finding the distance between two atoms whose positions are given in spherical polar coordinates $(r, \theta, \Phi)$. We first convert these spherical coordinates to Cartesian coordinates $(x, y, z)$ using the standard transformations:

• $x = r \sin \theta \cos \Phi$
• $y = r \sin \theta \sin \Phi$
• $z = r \cos \theta$

Converting Coordinates to Cartesian

Atom 1: $(r_1, \theta_1, \Phi_1) = (1, \frac{\pi}{2}, \frac{\pi}{2})$

• $x_1 = 1 \cdot \sin(\frac{\pi}{2}) \cos(\frac{\pi}{2}) = 1 \cdot 1 \cdot 0 = 0$
• $y_1 = 1 \cdot \sin(\frac{\pi}{2}) \sin(\frac{\pi}{2}) = 1 \cdot 1 \cdot 1 = 1$
• $z_1 = 1 \cdot \cos(\frac{\pi}{2}) = 1 \cdot 0 = 0$

Cartesian coordinates for Atom 1 are $(0, 1, 0)$.

Atom 2: $(r_2, \theta_2, \Phi_2) = (1, \frac{\pi}{4}, \frac{3\pi}{2})$

• $x_2 = 1 \cdot \sin(\frac{\pi}{4}) \cos(\frac{3\pi}{2}) = 1 \cdot \frac{\sqrt{2}}{2} \cdot 0 = 0$
• $y_2 = 1 \cdot \sin(\frac{\pi}{4}) \sin(\frac{3\pi}{2}) = 1 \cdot \frac{\sqrt{2}}{2} \cdot (-1) = -\frac{\sqrt{2}}{2}$
• $z_2 = 1 \cdot \cos(\frac{\pi}{4}) = 1 \cdot \frac{\sqrt{2}}{2} = \frac{\sqrt{2}}{2}$

Cartesian coordinates for Atom 2 are $(0, -\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2})$.

Calculating Interatomic Distance

The distance $d$ between two points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ in Cartesian coordinates is given by:

$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}$

Substituting the Cartesian coordinates:

$d = \sqrt{(0 - 0)^2 + (-\frac{\sqrt{2}}{2} - 1)^2 + (\frac{\sqrt{2}}{2} - 0)^2}$

$d = \sqrt{0^2 + (-( \frac{\sqrt{2}}{2} + 1))^2 + (\frac{\sqrt{2}}{2})^2}$

$d = \sqrt{(\frac{\sqrt{2}}{2} + 1)^2 + (\frac{\sqrt{2}}{2})^2}$

$d = \sqrt{(\frac{2}{4} + 2 \cdot \frac{\sqrt{2}}{2} \cdot 1 + 1^2) + \frac{2}{4}}$

$d = \sqrt{(\frac{1}{2} + \sqrt{2} + 1) + \frac{1}{2}}$

$d = \sqrt{1 + \sqrt{2} + 1} = \sqrt{2 + \sqrt{2}}$

Final Distance Calculation and Rounding

Now, we calculate the numerical value:

• $\sqrt{2} \approx 1.41421$
• $d = \sqrt{2 + 1.41421} = \sqrt{3.41421}$
• $d \approx 1.84776$ Å

Rounding the interatomic distance to two decimal places gives $1.85$ Å.