Answer
a. n=5
b. -0.544eV
c. $2.58\times10^{-34}J \cdot s $
d. $m_{\mathcal{l}}= -2,-1,0,1,2$
Work Step by Step
a. The principal quantum number is n=5.
b. Find the energy of the state using equation 27–15b.
$$E_5=-\frac{-13.6eV}{5^2}=-0.544eV$$
c. The “d” subshell has $\mathcal{l}=2$.
The magnitude of the angular momentum depends only on $\mathcal{l}$.
$$L=\sqrt{\mathcal{l}(\mathcal{l}+1)}\hbar=\sqrt{6}\hbar$$
$$=\sqrt{6}\hbar=\sqrt{6}(1.055\times10^{-34}J \cdot s) =2.58\times10^{-34}J \cdot s $$
d. The values of $m_{\mathcal{l}}$ range from $-\mathcal{l}$ to $\mathcal{l}$. Thus, for $\mathcal{l}=2$, $m_{\mathcal{l}}= -2,-1,0,1,2$.
