# Solutions 15

- Page ID
- 47396

## S15.1

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

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

m_{l} can be: -l, -l+1, -l+2, ..., 0, ..., l-1, l

m_{s} 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

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

this leads us to

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

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 E_{h} is the Hartree energy:

\[E_n = \dfrac{Z^2 E_h}{2 n^2}\]

$$E_{h}=m_{e} (\dfrac{e^2}{2 h \epsilon _{0} })^2=27.211 eV$$

- \(E_2 = \dfrac{(1)^2 (27.211\;eV)}{2 (2)^2} = 3.40\;eV\)
- \(E_4 = \dfrac{(1)^2 (27.211\;eV)}{2 (3)^2} = 0.850\;eV\)
- \(E_5 = \dfrac{(1)^2 (27.211\;eV)}{2 (4)^2} = 0.544\;eV\)
- \(E_6 = \dfrac{(1)^2 (27.211\;eV)}{2 (5)^2} = 0.378\;eV\)

### S15.4

Similar to 15.3, we are dealing with single electron system, but with Z replaced with 3 for Li and 4 for Be and for ground state n=1.

\[E_n = \dfrac{Z^2 E_h}{2 n^2}\]

Li: \(E_1 = \dfrac{(3)^2 (27.211\;eV)}{2 (1)^2} = 122.450\;eV=1.96 \times 10^{-17} J\)

Be: \(E_1 = \dfrac{(4)^2 (27.211\;eV)}{2 (1)^2} = 217.688\;eV=3.48 \times 10^{-17} J\)

## S15.5

This is a single electron case, so use the same formula as 15.4.

First calculate the energy difference between 3s and 2p levels:

$$ \Delta E=E_{2}-E_{3}=\dfrac{Z^2 E_{h}}{2} (\dfrac{1}{2^2}-\dfrac{1}{3^2})=\dfrac{6^2 \times 27.211 eV}{2} \times (\dfrac{1}{4}-\dfrac{1}{9})=68.0275 eV \times \dfrac{1.6 \times 10^{-19}}{1 eV}=1.088 \times 10^{-17} J $$

Then calculate the wavelength of light emitted when the electron transits between the two energy levels:

$$ \Delta E=h \nu =\dfrac{h c}{\lambda}$$

$$\lambda =\dfrac{h c}{\Delta E}=\dfrac{6.63 \times 10^{-34} Js \times 3 \times 10^8 m/s}{1.088 \times 10^{-17} J}=1.83 \times 10^{-8} m =18.3 nm $$

## S15.6

To calculate the probability of a 1s hydrogen electron being found within distance 2a_{0} of nucleus, we just need to find out the volume integral of the square of the absolute value of the wave function in that region. Assume the distance of the electron to the nucleus is r, the angle between r and z axis is \(\theta\) and the angle between x axis and the projection of r on the xy plane is \(\phi\). The 1s wavefunction is then

$$\Psi=\dfrac{1}{\sqrt{\pi}}(\dfrac{Z}{a_{0}})^{3/2} e^{-r}$$

Therefore,

$$P=\int_0^{2\pi} \! \int_{-\dfrac{\pi}{2}}^{\dfrac{\pi}{2}} \! \int_0^{2 a_0} | \Psi |^2 r^2 sin\theta\,dr\,d\theta\,d\phi. $$

## 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, m_{l}=-1, 0, 1 which correspond to 2P_{x}, 2P_{y}, 2P_{z}. 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, m_{l}, m_{s})

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

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

(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