Answer
See the detailed answer below.
Work Step by Step
First, we need to find the first three states of the hydrogen atom's angular momenta.
Recalling that the angular momentum is given by
$$L=mvr$$
Thus, the angular momentum of the hydrogen atom in terms of Plank’s constant $\hbar$, where $\hbar=h/2\pi$
$$\dfrac{L}{\hbar}=\dfrac{mv_nr_n}{\hbar}$$
$$\dfrac{L}{\hbar}=\dfrac{2\pi mv_nr_n}{h}$$
Plugging the known using Table 38.2:
For $n=1$
$$\dfrac{L_1}{\hbar}=\dfrac{2\pi(9.11\times 10^{-31})(2.19\times 10^6)(0.053\times 10^{-9})}{(6.63\times 10^{-34})}=\bf 1.002$$
Therefore,
$$\boxed{L_1=\hbar}$$
For $n=2$
$$\dfrac{L_2}{\hbar}=\dfrac{2\pi(9.11\times 10^{-31})(1.09\times 10^6)(0.212\times 10^{-9})}{(6.63\times 10^{-34})}=\bf 1.99$$
$$\dfrac{L_2}{\hbar}\approx \bf 2.0$$
Therefore,
$$\boxed{L_2=2\hbar}$$
For $n=3$
$$\dfrac{L_3}{\hbar}=\dfrac{2\pi(9.11\times 10^{-31})(0.73\times 10^6)(0.476\times 10^{-9})}{(6.63\times 10^{-34})}=\bf 2.999$$
$$\dfrac{L_3}{\hbar}\approx \bf 3.0$$
Therefore,
$$\boxed{L_3=3\hbar}$$
