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).

Question

An electron is confined to move in a one-dimensional box of length 1˚ A. Its energy in the first excited state is approximately

Answer 1

Instead of converting both points fully into Cartesian coordinates, this method finds the angle between the two position vectors directly using the spherical law of cosines, then applies the law of cosines for the two known radii.

Angle between the vectors: for two points at $(r_1,\theta_1,\Phi_1)$ and $(r_2,\theta_2,\Phi_2)$, the angle $\gamma$ between the vectors from the origin satisfies $\cos\gamma = \cos\theta_1\cos\theta_2 + \sin\theta_1\sin\theta_2\cos(\Phi_1-\Phi_2)$.
Plug in the given angles: $\theta_1=\pi/2$, $\theta_2=\pi/4$, $\Phi_1=\pi/2$, $\Phi_2=3\pi/2$. Then $\cos\theta_1=0$, $\cos\theta_2=\sin\theta_2=0.7071$, $\sin\theta_1=1$, and $\Phi_1-\Phi_2=-\pi$ so $\cos(\Phi_1-\Phi_2)=-1$.
Compute $\cos\gamma$: $\cos\gamma = (0)(0.7071) + (1)(0.7071)(-1) = -0.7071$, which is $\cos 135^\circ$, so the two position vectors are $135^\circ$ apart.
Law of cosines for the distance: with $r_1=r_2=1\ \overset{\circ}{\mathrm{A}}$, $d^2 = r_1^2+r_2^2-2r_1r_2\cos\gamma = 1+1-2(1)(1)(-0.7071) = 2+1.4142 = 3.4142$.

Taking the square root, $d = \sqrt{3.4142} = 1.8478\ \overset{\circ}{\mathrm{A}}$, which agrees with a direct Cartesian conversion of both points.

Let's summarize:

The spherical law of cosines gives the angle between two direction vectors straight from their $\theta$ and $\Phi$ values, no need to write out $x,y,z$.
Once the angle is known, the ordinary law of cosines on the triangle formed by the two radii and the connecting segment gives the distance.

So the interatomic distance, rounded to two decimal places, is $\boxed{1.85\ \overset{\circ}{\mathrm{A}}}$.

Answer 2

Instead of converting both points fully into Cartesian coordinates, this method finds the angle between the two position vectors directly using the spherical law of cosines, then applies the law of cosines for the two known radii.

Angle between the vectors: for two points at $(r_1,\theta_1,\Phi_1)$ and $(r_2,\theta_2,\Phi_2)$, the angle $\gamma$ between the vectors from the origin satisfies $\cos\gamma = \cos\theta_1\cos\theta_2 + \sin\theta_1\sin\theta_2\cos(\Phi_1-\Phi_2)$.
Plug in the given angles: $\theta_1=\pi/2$, $\theta_2=\pi/4$, $\Phi_1=\pi/2$, $\Phi_2=3\pi/2$. Then $\cos\theta_1=0$, $\cos\theta_2=\sin\theta_2=0.7071$, $\sin\theta_1=1$, and $\Phi_1-\Phi_2=-\pi$ so $\cos(\Phi_1-\Phi_2)=-1$.
Compute $\cos\gamma$: $\cos\gamma = (0)(0.7071) + (1)(0.7071)(-1) = -0.7071$, which is $\cos 135^\circ$, so the two position vectors are $135^\circ$ apart.
Law of cosines for the distance: with $r_1=r_2=1\ \overset{\circ}{\mathrm{A}}$, $d^2 = r_1^2+r_2^2-2r_1r_2\cos\gamma = 1+1-2(1)(1)(-0.7071) = 2+1.4142 = 3.4142$.

Taking the square root, $d = \sqrt{3.4142} = 1.8478\ \overset{\circ}{\mathrm{A}}$, which agrees with a direct Cartesian conversion of both points.

Let's summarize:

The spherical law of cosines gives the angle between two direction vectors straight from their $\theta$ and $\Phi$ values, no need to write out $x,y,z$.
Once the angle is known, the ordinary law of cosines on the triangle formed by the two radii and the connecting segment gives the distance.

So the interatomic distance, rounded to two decimal places, is $\boxed{1.85\ \overset{\circ}{\mathrm{A}}}$.

Show Hint

Correct Answer: 1.85

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