HW 1 - Review of QM

1.5,6,9

1.5

Theorem

Describe the quantum numbers of all orbitals in the $4 f$ subshell.

We know that there are 7 orbitals in any $f$ subshell. This is because there are up to 14 electrons in that subshell. We are given $4 f$ specifically, which indicates that $n = 4$ is our principal quantum number, and $l = 3$ associated with the $f$ subshell.

Note for each of these, the magnetic quantum number $m_{l}$ takes on values $- l, . . ., l$ all integer values. They represent the orbitals in a subshell which actually contain electrons. Hence:

Equivalent Orbital Configuration	$n$	$l$	$m_{l}$
$4 f^{2}$	4	3	-3
$4 f^{4}$	4	3	-2
$4 f^{6}$	4	3	-1
$4 f^{8}$	4	3	0
$4 f^{10}$	4	3	1
$4 f^{12}$	4	3	2
$4 f^{14}$	4	3	3

1.6

Theorem

Describe the ground state electron configuration for Silicon.

Silicon (Si) has an atomic number 14, hence it can be represented as:

1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{2} \equiv [Ne] 3 s^{2} 3 p^{2}

1.9

Theorem

Light of wavelength $500 nm$ is incident on a double-slit with a distance $0.01 mm$ between the slits. Calculate the angle made by the first order bright (constructive interference) fringe $(n = 1)$ . Use $λ = \frac{d \sin (θ)}{n}$

We are given $d$ and $λ$ and $n = 1$ , so then solve for $θ$ :

\Rightarrow \sin (θ) = \frac{n λ}{d} = \frac{1 \cdot 500 nm}{0.01 mm} = 0.05 \Rightarrow θ = \arcsin (0.05) = 0.0500 \approx 2.866 °

Bonus Problems (Binary)

1

Convert the following numbers into binary: 18, 42, 89, 127.

18_{10} = 16 + 2 = 2^{4} + 2^{1} = 1 \cdot 2^{4} + 0 \cdot 2^{3} + 0 \cdot 2^{2} + 1 \cdot 2^{1} + 0 \cdot 2^{0} = 10010_{2}

42_{10} = 32 + 8 + 2 = 1 \cdot 2^{5} + 0 \cdot 2^{4} + 1 \cdot 2^{3} + 0 \cdot 2^{2} + 1 \cdot 2^{1} + 0 \cdot 2^{0} = 101010_{2}

89_{10} = 64 + 16 + 8 + 1 = 1 \cdot 2^{6} + 0 \cdot 2^{5} + 1 \cdot 2^{4} + 1 \cdot 2^{3} + 0 \cdot 2^{2} + 0 \cdot 2^{1} + 1 \cdot 2^{0} = 1011001_{2}

127_{10} = 64 + 32 + 16 + 8 + 4 + 2 + 1 = 1111111_{2}

2

Add the following pairs of binary numbers, then convert the result to base 10:

101 + 110
1101 + 1010
10011 + 11000
1111001 + 1011100
1101101 + 1010101