Question

The positions of two atoms in spherical polar coordinates \((r, \theta, \Phi)\) are \((1, \frac{\pi}{2}, \frac{\pi}{2})\) and \((1, \frac{\pi}{4}, \frac{3\pi}{2})\), where the distance is in \(\overset{\circ}{\mathrm{A}}\) and the angles are in radian. The interatomic distance (in \(\overset{\circ}{\mathrm{A}}\)) is (rounded off to two decimal places).

Hint:

Convert each point to Cartesian using \(x=r\sin\theta\cos\Phi\), \(y=r\sin\theta\sin\Phi\), \(z=r\cos\theta\), then use the 3D distance formula (or the spherical law of cosines directly).

Answer 1

Correct Answer: 1.85

Step 1: Write down the spherical-to-Cartesian conversion.
For a point at \((r, \theta, \Phi)\) in spherical polar coordinates, the Cartesian coordinates are
\[ x = r\sin\theta\cos\Phi, \quad y = r\sin\theta\sin\Phi, \quad z = r\cos\theta \] Here \(\theta\) is measured from the \(z\)-axis and \(\Phi\) is the azimuthal angle in the \(xy\)-plane.

Step 2: Convert atom 1, \((1, \pi/2, \pi/2)\).
\(\sin(\pi/2)=1\), \(\cos(\pi/2)=0\).
\[ x_1 = 1(1)\cos(\pi/2) = 0, \quad y_1 = 1(1)\sin(\pi/2) = 1, \quad z_1 = 1(0) = 0 \] So atom 1 is at \((0, 1, 0)\).

Step 3: Convert atom 2, \((1, \pi/4, 3\pi/2)\).
\(\sin(\pi/4)=\cos(\pi/4)=\dfrac{1}{\sqrt2}\approx 0.7071\); \(\cos(3\pi/2)=0\), \(\sin(3\pi/2)=-1\).
\[ x_2 = 1(0.7071)(0) = 0, \quad y_2 = 1(0.7071)(-1) = -0.7071, \quad z_2 = 1(0.7071) = 0.7071 \] So atom 2 is at \((0, -0.7071, 0.7071)\).

Step 4: Apply the distance formula.
\[ d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2} \] \[ d = \sqrt{(0-0)^2 + (-0.7071-1)^2 + (0.7071-0)^2} \] \[ d = \sqrt{(-1.7071)^2 + (0.7071)^2} = \sqrt{2.9142 + 0.5000} = \sqrt{3.4142} \] \[ d \approx 1.8477\ \overset{\circ}{\mathrm{A}} \]
Final Answer:
Rounded to two decimal places, the interatomic distance is \[ \boxed{1.85\ \overset{\circ}{\mathrm{A}}} \]