Ch. 1 Structure of the Atom

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Atomic Discoveries Summarized

Name Symbol Charge (Coulombs) Mass
Electron e e or e e^- 1.602 1 0 19 - 1.602 \cdot 10^{- 19} (-1) 9.109 1 0 28 9.109 \cdot 10^{- 28} ( 5.486 1 0 4 5.486 \cdot 10^{- 4} )
Proton p p 1.602 1 0 19 1.602 \cdot 10^{- 19} (+1) 1.673 1 0 24 1.673 \cdot 10^{- 24} (1.0073)
Neutron n n 0 1.675 1 0 25 1.675 \cdot 10^{- 25} (1.0087)

Electromagnetic Spectrum


Light

( ( Wavelength ) ( )( Frequency ) = ( )=( Speed of Light ) )
λ v = c = 299 , 792 , 458  m/s 3.0 1 0 8  m/s \textcolor{#EE4266}{\lambda} \textcolor{#F79256}{v}=\textcolor{yellow}{c}=299,792,458\text{ m/s}\approx 3.0\cdot10^8\text{ m/s}
( ( Energy ) = ( )=( Planck's constant ) ( )( Frequency ) )
E = h v = h c λ = m c 2 , h = E=\textcolor{lightblue}{h}\textcolor{#F79256}{v}=\textcolor{lightblue}{h}\frac{\textcolor{yellow}{c}}{\textcolor{#EE4266}{\lambda}}=mc^2,\textcolor{lightblue}{h}= 6.63 1 0 34  J s 6.63 \cdot 10^{- 34} \text{ J} \cdot \text{s}

Example Problem

Find the frequency and energy of 400. nm light
λ = 400.  nm \lambda = 400 . \text{ nm}
( 400.  nm ) v = 3.0 1 0 8  m s 1 \left( 400 . \text{ nm} \right ) v = 3.0 \cdot 10^{8} \text{ m s}^{- 1}
v = 3.0 1 0 8  m s 1 4.00 1 0 7  m = 7.5 1 0 14  s 1 v = \frac{3.0 \cdot 10^{8} \text{ m s}^{- 1}}{4.00 \cdot 10^{- 7} \text{ m}} = 7.5 \cdot 10^{14} \text{ s}^{- 1}
E = h v = ( 6.63 1 0 34  J s ) ( 7.5 1 0 14  s 1 ) = 5.0 1 0 19  J E = h v = \left( 6.63 \cdot 10^{- 34} \text{ J s} \right ) \left( 7.5 \cdot 10^{14} \text{ s}^{- 1} \right ) = 5.0 \cdot 10^{- 19} \text{ J}


Energy in a orbital (principal quantum number) n n : E n = 2 π 2 m e 4 n 2 h 2 = 2.178 1 0 18 n 2  J E_{n} = \frac{- 2 \pi^{2} m e^{4}}{n^{2} h^{2}} = \frac{- 2.178 \cdot 10^{- 18}}{n^{2}} \text{ J}


Energy Levels

Principal Energy Levels (Shells): 1, 2, 3, 4,...
Sublevels (Subshells): s, p, d, f
Orbitals: 1, 3, 5, 7 pairs of e e^- can fit in sublevel


Electrons

Electron Configuration: e.g. 1 s 2 2 s 2 2 p 6 3 s 2 3 p 6 4 s 2 3 d 10 1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 4 s^{2} 3 d^{10} \ldots or abbreviated

Electron Configuration for F e \ce{Fe}

F e = 1 s 2 2 s 2 2 p 6 3 s 2 3 p 6 4 s 2 3 d 6 \ce{Fe}=1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^6
F e = [ A r ] 4 s 2 3 d 6 \ce{Fe}=[\ce{Ar}] 4s^2 3d^6

Valence Electrons: Outside electrons; sum of s and p orbitals
Hund's Rule: p, d, and f orbitals must have one electron in each orbital before they can pair up