Answer
See answers.
Work Step by Step
a. Treat the Earth as a point mass that is orbiting around the sun. The angular momentum here is given by L=mvr, where r is the radius of the orbit. Use equation 28–3 to find the orbital quantum number.
$$L=mvr =\sqrt{\ell(\ell+1)}\hbar\approx \ell \hbar$$
The final approximation holds because $\ell$ is very large.
$$L=M_{earth}\frac{2\pi r}{T}r= \ell \hbar$$
$$\ell=\frac{ M_{earth}2\pi r^2}{\hbar T}$$
$$\ell=\frac{(5.98\times10^{24}kg)2\pi (1.496\times10^{11}m)^2}{(1.055\times10^{-34}J\cdot s) (3.156\times10^7 s)}$$
$$\ell=2.53\times10^{74}$$
b. There are $2\ell + 1$ values of the magnetic quantum number $m_{\ell}$ for each value of $\ell$, each of which corresponds to a different orientation of the plane of Earth’s orbit.
The number of orientations $2\ell + 1$.
$$N=2\ell + 1=2(2.5255\times10^{74})+1=5.05\times10^{74}$$
