Solutions 15

S15.1

n can be any integer number: 1, 2, 3, ...

l can only be: 0, 1, 2, ... , (n-1)

ml can be: -l, -l+1, -l+2, ..., 0, ..., l-1, l

ms can only be: -1/2 or 1/2

S15.2

Identify the quantum numbers for the electron of interest (in our case, $$n=2$$; $$l =1$$)

Energy of the electron can be defined as

$E_n = \dfrac{-m_ee^4}{8n^2\epsilon_o^2h^2}$

$E_2 = \dfrac{-m_ee^4}{32\epsilon_o^2h^2}$

we have two possible wave functions

$\Psi_{210}= \dfrac{1}{\sqrt{32}} (\dfrac{z}{a_o})^{3/2}\sigma/e^{-\sigma/2} \cos{\theta}$

and

$\Psi_{21\pm1}= \dfrac{1}{\sqrt{32}} (\dfrac{z}{a_o})^{3/2}\sigma/e^{-\sigma/2} \sin{\theta}e^{\pm i\theta }$

The 2s electron has the same energy.

S15.3

The removal of an electron from a Hydrogen atom is akin to moving an electron from energy state n to a distance infinity. The energy required for this is called the ionization energy which is equivalent (with the possible exception of a sign change depending on how ionization energy is defined) to the energy of a one-electron atom like the Hydrogen atom. Thus, solve for the desired energy using the following equation for the energy of a one-electron atom where Eh is the Hartree energy:

$E_n = \dfrac{Z^2 E_h}{2 n^2}$

S15.7

(a) Each orbital can only contain at most 2 electrons, so (x) 2, (y) 2, (z) 2.

(b) When n=2, l=1 or 0. l=0 corresponds to 2s which is type x; when l=1, ml=-1, 0, 1 which correspond to 2Px, 2Py, 2Pz. There will be no d orbitals. Therefore, (x) 1, (y) 3, (z) 0.

(c) x: 4s orbital can only have 4, 0, 0, 1/2 (n, l, ml, ms)

y:4p orbital, l=1, ml has three possible values(-1, 0, 1), let us pick ml=1, we have: 2, 1, 0, 1/2 (n, l, ml, ms)

z. 4d orbital, l=2, ml can be (-2, -1, 0, 1, 2), let us pick ml=0, then we have: 3, 2, 0, 1/2 (n, l, ml, ms)

(d) x: 1

y: 2

z: 3

(e) x: l = 0, ml = 0

y: l = 1, ml = –1, 0, or +1

z: l = 2, ml = –2, –1, 0, +1, +2