Answer
See explanation.
Work Step by Step
The possible values of the angular momentum are limited by n.
For the N shell, n = 4.
a. The smallest $\mathcal{l}$ is zero, so the smallest orbital angular momentum is $\sqrt{l(l+1)}\hbar=0$.
b. The largest $\mathcal{l}$ is n-1, so the largest orbital angular momentum is $\sqrt{3(3+1)}\hbar=\sqrt{12}\hbar$. In SI units, that is $3.65\times10^{-34}kg \cdot m^2/s$.
c. The largest projection of the orbital angular momentum along any direction, say, the z-axis, is when m has its maximum value; here it is 3. The largest orbital angular momentum is $3\hbar $. In SI units, that is $3.16\times10^{-34}kg \cdot m^2/s$.
d. The largest spin angular momentum in a chosen direction is given by equation 41.36, $\frac{\hbar}{2}$. In SI units, that is $5.27\times10^{-35}kg \cdot m^2/s$.
e. The ratio is one-sixth.
