Materi Pelajaran Fisika SMA Semester 1/2 Kelas 12 Bab 8 Fisika Atom

membuat ulasan tentang perkembangan teori atom

Setelah mempelajari bab ini, Anda harus mampu:

menganalisis berbagai besaran fisis pada gejala kuantum dan batas-batas berlakunya relativitas Einstein dalam paradigma Fisika Modern.

Hasil yang harus Anda capai:

Fisika Atom

A. Evolusi Model

Atom

B. Atom Berelektron

Banyak

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Bab

8

Sumber: Chemistry: Matter and Its Changes, 2003 Atom terdiri atas proton dan neutron yang terletak pada inti serta elektron yang mengorbit pada kulit.

A. Evolusi Model Atom

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1. Model Atom Demokritus

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)$=5>75=21>7;1>@5>41@1D,;*077;945>71>=5=25B9>1=1,'%'+

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=1D5B91< 41> J1D I1>7 141 49 1<1= 9>9 49CECE> ?<58 21791> D5B;539< I1>7 D941; 41@1D 492179 <179 I19DE 1D?=

'141 =1C1 D5BC52ED 21>I1; @1B1 69<CE6 I1>7 =5>75=E;1;1> @5>41@1D>I1 =5>75>19 1D?= B7E=5>D1C9 I1>7 =5B5;1 7E>1;1> 2E;1> 4941C1B;1>@14181<@5B3?211>=5<19>;1>25B41C1B;1><?79;1-1<1E@E> @5>5=E1>=5B5;19>9C521D1CD5?B9D5D1@981<9>9D5<18=5>:149D?>771; @5>D9>7 E>DE; @5;5=21>71> D5?B9 69C9;1 1D?= 41> 69C9;1 9>D9 )59B9>7 45>71> @5B;5=21>71> J1=1> 21>I1; 69C9;1G1> 41> ;9=91G1> I1>7 D5BD1B9; E>DE; =5=@5<1:1B9 D5?B9 1D?= 9>9 D5BEC =5=2E1D 5F?<EC9 =?45< 1D?= C589>771 4941@1D;1> =?45< 1D?= C5@5BD9 C5;1B1>7 9>9

2. Model Atom Dalton

6/5 (3:65M141<18C5?B1>79<=EG1> >77B9CI1>7;1<9

@5BD1=1 =5>I1D1;1> D5?B9 D5>D1>7 1D?= 25B41C1B;1> 81C9< 5;C@5B9=5> I1>7 49<1;E;1>69,7/ !86;9: C5?B1>7 ;9=91G1> 41B9 'B1>39C I1>7

=5>I1D1;1> 218G1 C5>I1G1 49CECE> ?<58 E>CEBE>CEB 45>71> @5B21>49>71> D5D1@ 1C9< 5;C@5B9=5>!86;9: 9>9 49C52ED E;E=

'5B21>49>71> *5D1@ '141 !86;9: =5<1;E;1> 5;C@5B9=5> 45>71>

=5B51;C9;1> E>CEB 894B?75> 41> E>CEB ?;C975> 1 =5>5=E;1> C5>I1G1 19B 45>71> @5B21>49>71> D5D1@ I19DE 71B >41 <5298 :5<1C @5B81D9;1> D125< 25B9;ED

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Tes Kompetensi Awal

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Tokoh

John Dalton (1766–1844)

John Dalton adalah seorang ilmuwan Inggris. Dia orang pertama yang mengemukakan tentang teori atom berdasarkan hasil eksperimen yang dilakukan Joseph Proust. Dalton menyatakan bahwa atom adalah bagian terkecil dari benda yang tidak dapat dibagi-bagi lagi.

Sumber: www.allbiographies.com

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3. Model Atom Thomson

'141 1;89B 1214 . .$/64(9 3<( +0965 25B81C9< =5>5=E;1>

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'141C5?B1>769C9;1G1> >77B9C25B>1=169,7/6/5$/64965

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(4)(8 25B9;ED

Tokoh

Sir J.J. Thomson (1856–1940)

Thomson adalah seorang ahli Fisika yang lahir di Checham Hill, Greater Manchester, Inggris. Dia belajar di Cambridge dan meraih gelar profesor pada 1884. Dia meraih hadiah Nobel Fisika tahun 1906. Eksperimennya yang terkenal tentang tabung sinar katode. Dari eksperimennya, dia menemukan nilai nisbah e/m.

Sumber: www.allbiographies.com

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Gambar 8.1

Percobaan Thomson

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katode anode

layar fluoresensi

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B

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B F

v

Jika ada elektron yang bergerak dengan kecepatan v dalam suatu medan magnet B, arah gaya Lorentznya mengikuti kaidah tangan kanan.

pengamat mikroskopis

minyak disemprot

lubang kecil radiasi sinar-X

pelat negatif

tetes minyak

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Sumber: Chemistry, 2001

Gambar 8.2

Model Atom Thomson

Gambar 8.3

Percobaan tetes minyak Millikan

– –

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F_S F_A

mg

Gambar 8.4

Gaya berat, gaya Archimedes, dan gaya stokes bekerja pada tetesan minyak pada percobaan Millikan.

Gambar 8.5

Tetesan minyak berada dalam kesetimbangan karena adanya gaya berat, gaya Archimedes, dan gaya elektrostatis.

F_{elektrostatis} F_A

mg

Tugas Anda 8.1

Dengan menggunakan logika dan pemahaman Anda tentang konsep gerak lurus, apakah Anda dapat menduga bagaimana Millikan mengukur kecepatan (v) dari tetesan minyak tersebut? Jelaskan dengan alasan-alasan ilmiah menurut pendapat Anda.

*

%-

M _{* .}

*

%

-.

M

)E2CD9DEC9;1>!,89(4((5@ ;5!,89(4((5C589>7714941@1D;1>

)

%

-.

.

M

&<58 ;1B5>1_{- %}!,89(4((5 @ 41@1D 49DE<9C;1> =5>:149

)

_%

. .

M

)5D5<18 =5>7E<1>7 5;C@5B9=5> D5BC52ED 2525B1@1 ;1<903302(5

=5>41@1D;1> ;5C9=@E<1> C521719 25B9;ED

5C1B>I1 =E1D1> D5D5C1> =9>I1; D941; @5B>18 25B>9<19 <5298 ;539<

41B9PM

$E1D1>D5D5C1>>I1C5<1<E;5<9@1D1>2E<1D41B9PM

1B9 ;5C9=@E<1> 9>9 4941@1D;1> 81C9< 218G1 =E1D1> 5<5;DB?> 141<18

PM_PM

41> =1CC1>I1 141<18

P ;7

%

%

P

P

M_;7

4. Model Atom Rutherford

*5?B9 1D?= *8?=C?> @141 49E:9 ?<58 C5?B1>7 69C9;1G1> 1C1< >77B9C85,9:";:/,8-68+ 1=5>7E:9;525>1B1>D5?B9*8?=C?>45>71>

=5<1;E;1>@5B3?211>=5>77E>1;1>@1BD9;5<1<61I1>749D5=21;;1>@141 C52E18;5@9>7<?71=5=1CI1>7C1>71DD9@9C'1BD9;5<1<61141<18@1BD9;5< I1>7 49@1>31B;1> ?<58 E>CEB B149?1;D96 25B=E1D1> <9CDB9; @?C9D96 I1>7 25C1B>I14E1;1<9=E1D1>5<5;DB?>41>=1CC1>I15=@1D;1<9=1CC1@B?D?> );5=1 @5B3?211> (ED85B6?B441@1D >41 <981D @141(4)(8

sumber partikel alfa

beberapa partikel alfa dipantulkan kembali

keping tipis emas

pancaran alfa

sebagian partikel alfa dibelokkan

sebagian partikel alfa menembus keping logam

layar detektor

Gambar 8.6

Percobaan Rutherford

Tokoh

Ernest Rutherford (1871–1937)

Ernest Rutherford lahir di New Zealand pada 1871. Ia mendapat beasiswa di Universitas Cambridge, Inggris pada 1894. Di sana, ia mulai melakukan penelitian tentang radiasi elektromagnetik, tetapi kemudian perhatiannya beralih ke sinar-X dan radioaktivitas. Ia menunjukkan bahwa ada perbedaan tipe radioaktivitas, yang ia namakan radiasi alfa, beta, dan gamma. Ia mendapat hadiah Nobel untuk bidang kimia pada 1908. Pada 1911, Rutherford menjadi seorang profesor di Universitas Manchester, ia menemukan bahwa hampir seluruh massa atom berkumpul di inti atom. Sumber: Science Encyclopedia, 1998

partikel alfa

inti atom

menembus dibelokkan

dipantulkan

";:/,8-68+ =5=9<9;9 1CE=C9 218G1 :9;1 D5?B9 1D?= *8?=C?> 25>1B

C5<EBE8 @1BD9;5< 1<61 45>71> 5>5B79 I1>7 25C1B 81BEC =5>5=2EC <EBEC ;5@9>7 D9@9C 5=1C D5BC52ED ;1B5>1 1D?=1D?= ;5@9>7 <?71= 5=1C >5DB1< D941; =5>781<1>79 @1BD9;5< 1<61 I1>7 25B=E1D1> <9CDB9; @?C9D96 )5D5<18 49<1;E;1> 2525B1@1 ;1<9 @5B3?211> 49C9=@E<;1> 218G1 C521791> 25C1B @1BD9;5< 1<61 =5>5=2EC ;5@9>7 <?71= D9@9C <EBEC =5>75>19 <1I1B ;1> D5D1@9 2525B1@1 @1BD9;5< 1<61 I1>7 <19>>I1 4925<?;;1> 218;1> 141 I1>7 49@1>DE<;1>)5D91@CE4ED@5=25<?;1>41>@5=1>DE<1>C9>1B1<61D5BC52ED 49E;EB 45>71> D5<9D9 ?<58";:/,8-68+

1;D1 9>9 ;5=E491> 491>1<9C9C ?<58";:/,8-68+ C521719 25B9;ED

5B41C1B;1>E;E=?E<?=2@1BD9;5<1<61I1>725B=E1D1>@?C9D9681>I1 1;1> 4925<?;;1> 1D1E 49@1>DE<;1> ?<58 CE1DE =E1D1> C5:5>9C I19DE =E1D1>@?C9D96BD9>I141<1=1D?=<?71=5=1C81BEC141=E1D1><9CDB9; @?C9D96 41> D941; D5BC521B 49 C5<EBE8 1D?= =5<19>;1> D5B@EC1D 49 CE1DE D5=@1D C589>771 =5>?<1; @1BD9;5< 1<61 D1C 41C1B 61;D1 9>9";:/,8-68+

25B;5C9=@E<1> C521719 25B9;ED

1 $E1D1><9CDB9;1D?=41>C521791>25C1B=1CC11D?=D5B@EC1D49CE1DE D9D9; I1>7 49C52ED ?<58";:/,8-68+ C521719 9>D9 1D?= >D9 1D?= 9>9

=5BE@1;1> CE1DE 415B18 I1>7 C1>71D ;539< 45>71> 491=5D5B C5;9D1B M₌

2 '141:1B1;I1>7B5<1D96:1E841B99>D91D?=D5BC52ED@1BD9;5<25B=E1D1> >571D961D1E5<5;DB?>25B541B=5>75<9<9>799>D9$E1D1><9CDB9;@?C9D96 I1>725B@EC1D499>D91D?=>9<19>I1C1=125C1B45>71>:E=<18=E1D1> <9CDB9; >571D96 41B9 5<5;DB?>5<5;DB?> I1>7 =5>75<9<9>79>I1

$5C;9@E><5298219;41B9@141D5?B9!!*8?=C?>>1=E>D5?B91D?= (ED85B6?B4 =1C98 =5>I9=@1> ;5<5=181> "5<5=181> D5?B9 1D?= (ED85B6?B4 141<18 C521719 25B9;ED

*5?B9 1D?= (ED85B6?B4 D941; 41@1D =5>:5<1C;1> C@5;DBE= 3181I1 I1>7 49@1>31B;1> ?<58 1D?= 894B?75> ;5D9;1 71C 894B?75> D5BC52ED 49@1>1C;1> 1D1E 49=1CE;;1> ;5 41<1= D12E>7 41> 4925B9 2541 @?D5>C91< <9CDB9; I1>7 D9>779 1>D1B1 E:E>7E:E>7 D12E>7 D5BC52ED *5?B9 1D?= (ED85B6?B4 D941; 41@1D =5>:5<1C;1> ;5CD129<1> 1D?=

5B41C1B;1> E;E= ?E<?=2 5<5;DB?> I1>7 25B9>D5B1;C9 45>71> 9>D91D?=1;1>=5>71<1=971I1?E<?=2I1>7:E7125B6E>7C9C521719 71I1 C5>DB9@5D1< ;921D>I1 5<5;DB?> =5>71<1=9 @5B35@1D1> @5B35@1D1> C5>DB9@5D1< $5>EBED D5?B9 75<?=21>7 5<5;DB?=17>5D9; I1>749;5=E;1;1>?<58(>=,33:9;1=E1D1>5<5;DB?>=5>71<1=9

@5B35@1D1> =E1D1> D5BC52ED 1;1> =5=1>31B;1> 75<?=21>7 5<5;DB?=17>5D9; !9;1 45=9;91> 5>5B79 5<5;DB?> 25B;EB1>7 41> 1;89B>I1 1;1> :1DE8 D5BD1B9; ;5 9>D9 1D?=

5. Spektrum Atom Hidrogen

)E1DE71CI1>749=1CE;;1>;541<1=D12E>741>;54E1E:E>7D12E>7 D5BC52ED 4925B9 2541 @?D5>C91< <9CDB9; I1>7 D9>779 1;1> =5=1>31B;1> 3181I1 "5D9;1 71C 894B?75> 49=1CE;;1> 41<1= D12E>7 @5<E3ED1> 71C ;5=E491>49<1<E9?<58C9>1B;1D?45=1;171C894B?75>1;1>=5=1>31B;1> 3181I1 181I1 9>9 ;5=E491> 491>1<9C9C ?<58+(#,*'%,* I19DE C52E18

1<1D E>DE; =5<981D 41> =5=?DB5D C@5;DBE= 3181I1 C5;1<97EC =5>7E;EB @1>:1>775<?=21>7>I15B9;ED9>9141<18C@5;DBE=3181I1@141415B18 D1=@1; I1>7 49@1>31B;1> ?<58 71C 894B?75>

inti atom

elektron

Gambar 8.7

Skema penyimpangan sinar alfa.

Gambar 8.8

)5?B1>7 7EBE C5;?<18 =5>5>718 25B;521>7C11> )G54916/(55,9 (34,8 =5>31D1D 218G1 ;5<?=@?; @1>:1>7 75<?=21>7 C@5;DBE= 71B9C

@1>31B 894B?75> 41<1= 415B18 D1=@1; 41@1D 493?3?;;1> C531B1 D5@1D 45>71> BE=EC C521719 25B9;ED

&

# &

M

Keterangan:

& 29<1>71>I1>725B>9<19=E<1941B9 # ;?>CD1>D1>=

@1>:1>7 75<?=21>7 >=

!,89(4((5 @ 49;5>1< C521719 BE=EC 1<=5B 41> 45B5D1> 71B9C

C@5;DBE= I1>7 3?3?; 45>71> @5BC1=11> D5BC52ED 49>1=1;1> 45B5D 1<=5B"?+),8. ;5=E491> =5>5=E;1> BE=EC C5BE@1 @141 E>CEB 1<;1<9

#9%1"41>C5B5D1<=5BE>DE;C@5;DBE=894B?75>49DE<9C41<1= 29<1>71> 75<?=21>7. 9<1>71> 75<?=21>7 E>DE; 45B5D 1<=5B

45B5D 3181I1 D1=@1;141<18

_45>71>

&

&

M

Keterangan:

;?>CD1>D1(I425B7P₌

'1>:1>775<?=21>7@14145B5DC@5;DBE=1D?=894B?75>49D1=@9<;1> 45>71> @5BC1=11> C521719 25B9;ED

5B5D #I=1> 45B5D E<DB1 E>7E

_45>71>

&

&

5B5D 1<=5B 45B5D 3181I1 D1=@1;

_45>71>

&

&

5B5D '1C385> 45B5D 9>6B1 =5B18

_45>71>

&

&

5B5D B13;5DD 45B5D 9>6B1 =5B18

_45>71>

&

&

5B5D '6E>4 45B5D 9>6B1 =5B18

_45>71>

&

&

warna

frekuensi (Hz)

merah biru ungu

4 × 1014 _{5 × 10}14 _{6 × 10}14 _{7 × 10}14 _{8 × 10}14

Gambar 8.9

Spektrum hidrogen pada daerah tampak

Tantangan

untuk Anda

Diskusikanlah bersama teman Anda, apakah Anda dan teman Anda dapat

menduga mengapa dalam eksperimennya Rutherford

menggunakan sinar dan lempeng emas?

Tantangan

untuk Anda

Hitunglah panjang gelombang terpendek dan terpanjang untuk deret Balmer, deret Lyman, deret Pascha, deret Bracket, dan deret Pfund.

'14145B5D#I=1>:9;1;?>CD1>D1(I425B7P _{=89DE>7@1>:1>7}

75<?=21>71D5B@1>:1>741>2D5B@5>45; !1G12

+>DE;45B5D#I=1>25B<1;E@5BC1=11>C52171925B9;ED

_45>71>

&

&

1 '1>:1>775<?=21>7D5B@1>:1>74981C9<;1>E>DE;&

_{= =}

PM₌

2 '1>:1>775<?=21>7D5B@5>45;4981C9<;1>E>DE;&K

_{= =}

K

=

Contoh

8.1

6. Model Atom Bohr

)5D5<18";:/,8-68+=5>75=E;1;1>218G1=1CC141>=E1D1>@?C9D96

1D?= D5B89=@E> @141 CE1DE 415B18 ;539< 49 @EC1D 1D?= 69C9;1G1> 1C1< 5>=1B;0,39 6/8 @141 =5>75=E;1;1> 218G1 CDBE;DEB 1D?=

=9B9@ 45>71> C9CD5= D1D1 CEBI1 >D9 1D?= 25B141 49 @EC1D C9CD5= 1D?= 41>5<5;DB?>25B141@141<9>D1C1>D5BD5>DEI1>749C52ED;E<9D1D?=+>DE; =5>7E1D;1>D5?B91D?=I1>749ECE<;1>>I16/8=5>IECE>4E1@?CDE<1D

41<9< D5>D1>7 1D?= '?CDE<1D @5BD1=1

1<1=CE1DEC9CD5=1D?=5<5;DB?>D941;=5=1>31B;1>B1491C9;5D9;1 91 =5>7?B29D 9>D9 1D?= #9>D1C1><9>D1C1> 5<5;DB?> =5>7?B29D D1>@1 =5=1>31B;1> B1491C9 49>1=1;1> <9>D1C1> CD1C9?>5B 1D1E ?B29D CD1C9?>5B '?CDE<1D 9>9 25B<1G1>1> 45>71> D5?B9 5<5;DB?=17>5D9; ;<1C9;

'?CDE<1D ;54E1

'5=1>31B1>41>@5>I5B1@1>75<?=21>75<5;DB?=17>5D9;41<1=CE1DE 1D?= 25B8E2E>71> 45>71> DB1>C9C9 5<5;DB?> 41B9 4E1 <9>D1C1> CD1C9?>5B

+>DE; C545B81>1>I1 @5B81D9;1>(4)(8 $?45< 1D?= ?8B

E>DE; 1D?= 894B?75> =5=@5B<981D;1> 218G1 C52E18 5<5;DB?> =5>79D1B9 C52E18 9>D9 1D?= 25B=E1D1> @?C9D96 !1B9:1B9 ?B29D <9>7;1B1>>I1* 41>

5<5;DB?> 25B=1CC1% 25B75B1; 45>71> <1:E <9>51B D5D1@. 1I1 D1B9;

?E<?=2 25B@5B1> =5=25B9;1> @5B35@1D1> C5>DB9@5D1< C589>771

_{3?E<?=2C5>DB9@5D1<}

) ) _{%. %.}

* *

* *

&<58 ;1B5>1 E>DE; 1D?= 894B?75> @5BC1=11> D5BC52ED =5>:149

* %.* %. *

>5B79;9>5D9;5<5;DB?>141<18 _&<58;1B5>1

#

%. %. #

*

=1;1

_#

* M

Gambar 8.10

Model atom Bohr (Z = 1) untuk hidrogen m

v

F

r

inti

elektron

41@E>5>5B79@?D5>C91<5<5;DB?>141<185>5B79@?D5>C91<?E<?=2I1>7 25C1B>I1 C521719 25B9;ED

( * _M

C589>771 5>5B79 D?D1< I1>7 49=9<9;9 ?<58 5<5;DB?> 141<18

# ( * * * _M "5D5B1>71>

PM _%=

=E1D1> <9CDB9; * :1B9:1B9 <9>D1C1> =

a. Jari-Jari Lintasan Stasioner

9C9;1 ;<1C9; =5B1=1<;1> 218G1 C52E18 =E1D1> 5<5;DB9; I1>7 =5>71<1=9@5B35@1D1>C5@5BD95<5;DB?>I1>7=5>7?B29D41<1==?45<1D?= 81BEC=5B1491C9;1>5>5B795<5;DB?=17>5D9;C531B1;?>D9>E"5D9;15>5B79 9>9 49@1>31B;1> 5>5B79 D?D1< 5<5;DB?> =5>EBE> 41> 5<5;DB?> 1;1> 25B75B1; C@9B1< =5>E:E 9>D9 1D?= C589>771 1D?= 1;89B>I1 :1DE8 ;5 9>D9 1D?= +>DE; =5>71D1C9 ;5CE<9D1> 9>96/8 =5>7ECE<;1> 7171C1> ?B29D

CD1C9?>5B@?CDE<1D@5BD1=11B9C9>96/8=5>I9=@E<;1>218G141<1=

;51411> 9>9 =?=5>DE= CE4ED ?B29D1< 5<5;DB?> 25B>9<19 ;5<9@1D1> 41B9 81B71 1D1E 49DE<9C 45>71>

_M

,5;D?B =?=5>DE= CE4ED 41<1= 69C9;1 ;<1C9; 494569>9C9;1> C521719

+>DE;=?=5>DE=CE4ED5<5;DB?>I1>725B541B=5>75<9<9>79

9>D9 1D?=* D571; <EBEC( C589>771 @5BC1=11> =?=5>DE= CE4ED 41@1D

49C545B81>1;1> =5>:149 *( %.* !149 =5>EBED @?CDE<1D ?8B

=?=5>DE= CE4ED 141<18

% . *& M

"5D5B1>71>

&29<1>71>2E<1D&

5>71> =5=1CE;;1> 81B71. @141!,89(4((5 @ ;5 41<1=

!,89(4((5 @ 49@5B?<58

# * %. *

% &%* *

_%*& *

* % M !9;1 %

4941@1D;1>*&

Tokoh

Niels Bohr (1885–1962)

Pada awal 1900-an, struktur atom menjadi lebih jelas, namun hukum Fisika pada saat itu tidak dapat menjelaskan mengapa elektron tidak cepat menghasilkan spiral menuju inti. Niels Bohr, murid

Rutherford, membantu memecah-kan misteri itu dengan menunjukmemecah-kan bahwa elektron-elektron hanya diberi energi tertentu. Ia menemukan bahwa elektron-elektron yang memiliki energi terendah mengorbit paling dekat ke inti, dan elektron dengan energi tinggi mengorbit paling jauh. Pada 1913, Niels Bohr juga menjelaskan hubungan antara materi dan cahaya. Ia menunjukkan bahwa jika elektron berpindah dari satu tingkat energi ke tingkat energi lainnya, elektron-elektron itu mengeluarkan atau menyerap paket radiasi dalam bentuk cahaya. Paket ini dinamakan foton.

1B9!,89(4((5 @ 41@1D 49@5B?<58 :1B9:1B9 5<5;DB?> I1>7

49@5B;5>1>;1> E>DE; C5D91@ <9>D1C1> 5C1B1>_? 494569>9C9;1> C521719

:1B9:1B9?8B+>DE;1D?=894B?75>6/8=5>41@1D;1>81B71_?

>= 1D1E 1>7CDB?= 5B41C1B;1> D5?B9 1D?= ?8B 9>9 :1B9:1B9 ?B29D 5<5;DB?>81>I141@1D25B>9<19_?????41>C5D5BEC>I1!1B9

:1B91D?==5>EBEDD5?B99>9D941;@5B>1825B>9<19_?1D1E_? b. Energi Elektron Berdasarkan Teori Bohr

5C1B 5>5B79 5<5;DB?> 41@1D 49@5B?<58 45>71> =5>7712E>7;1>

!,89(4((5 @ 41>!,89(4((5 @ I19DE

& %

&

M

'5BC1=11> 9>9 =5>I1D1;1> 218G1 5>5B79 5<5;DB?> D5B;E1>D9C1C9 BD9>I1 81>I1 >9<19>9<19 5>5B79 D5BD5>DE I1>7 49@5B;5>1>;1> !9;1 41B9 @5BC1=11> D5BC52ED C5=E1 >9<19 D5D1@>I1 4989DE>7 4941@1D @5BC1=11> I1>7 <5298 C545B81>1 C5@5BD9 25B9;ED

5, &

&

M

'141D9>7;1DD5B5>418>9<19&5>5B795<5;DB?>C525C1BM5,

41> 25B541B @141 :1B9:1B9 ?8B L *9>7;1D 5>5B79 5<5;DB?> @141 ;51411> I1>7 <5298 D9>779 49DE>:E;;1> @141(4)(8

'1415<5;DB?>41>9>D91D?=I1>7D5B@9C18C1>71D:1E8I19DEE>DE;

&49@5B?<58!9;15<5;DB?>9DE25BDB1>C9C91D1E25B@9>41841B9

<9>D1C1> <E1B ;5 <9>D1C1> 41<1= C589>771 91 25B141 @141 ?B29D D5BD5>DE 41<1= <9>D1C1>& D5BD5>DE 5<5;DB?> 1;1> =5=9<9;9 81B71 5>5B79 <5298

;539< 41B9@141 ;51411> C5=E<1 '141 ;51411> D5BC52ED 1;1>

49@1>31B;1> 5>5B79 C525C1B_> )521<9;>I1 :9;1 5<5;DB?> @141 ;51411> & 5<5;DB?>>I1 41@1D 492521C;1> 41B9 9;1D1> 9>D9 45>71> =5=25B9>I1

5>5B79 C525C1B_> >5B79 9>9 49;5>1< 45>71> 5>5B79 9;1D ;51411>& !9;1

5>5B79 I1>7 492521C;1> =5<52989 5>5B79 9;1D ;5<52981> 5>5B79 41@1D =E>3E< C521719 5>5B79 ;9>5D9; 5<5;DB?> I1>7 ;9>9 D5<18 2521C 41> D941; D5B9;1D <179 @141 9>D9 1D?=

?8B=5=@?CDE<1D;1>218G1=5C;9@E>5<5;DB?>D941;=5=1>31B;1> B1491C9 5<5;DB?=17>5D9; ;5D9;1 25B541B @141 CE1DE D9>7;1D D5BD5>DE 5<5;DB?>41@1D25B@9>41841B9CE1DE;51411>5>5B79D5BD5>DE;5;51411> 5>5B79 I1>7 <19> )5<9C98 5>5B79 1>D1B1 4E1 ;51411> D5BC52ED =E>3E< C521719 C52E18 B1491C9 75<?=21>7 5<5;DB?=17>5D9; 25B5>5B79 BD9>I1

:9;15<5;DB?>25B@9>41841B9;51411>1G1<;5;51411>1;89B1;1>D5B:149 @1>31B1> 1D1E @5>I5B1@1> B1491C9 5<5;DB?=17>5D9; B5;E5>C9 75<?=21>7 5<5;DB?=17>5D9; I1>7 49C5B1@ 1D1E 49@1>31B;1> 41B9 4E1 ;51411> CD1C9?>5B 141<18

!

M

"5D5B1>71>

_! 5>5B79 =E<1=E<1 !

;?>CD1>D1'<1>3;PM_!C

5>5B79 1;89B !

6B5;E5>C9 B1491C9 75<?=21>7 5<5;DB?=17>5D9; J Gambar 8.11

Tingkat-tingkat energi atom hidrogen.

n n = 4

n = 3

n = 2

E = –0,8 eV

E = –1,5 eV

E = –3,4 eV

n = 1 E = –13,6 eV

E = 0

Pembahasan Soal

Dalam model atom Bohr, energi yang dibutuhkan oleh elektron hidrogen untuk pindah dari orbit dengan bilangan kuantum 1 ke 3 adalah ....

a. 1,50 eV b. 1,90 eV c. 2,35 eV d. 12,10 eV e. 13,60 eV

UMPTN 1992 Pembahasan:

Energi yang dibutuhkan E = E₃ – E₁

Diketahui bahwa energi dasar = E₁

= –13,6 eV Maks,

E₃ = 1 2

E n

= 2

13,6 eV 3

Jadi,

E= 13,6 eV

9 – (– 13,6 eV)

= 12,10 eV

9DE>7<18;535@1D1>5>5B79;9>5D9;5>5B79@?D5>C91<41>5>5B79D?D1<5<5;DB?>@141 29<1>71>;E1>DE=&25B41C1B;1>D5?B9?8BE>DE;1D?=894B?75>

(=()

9;5D18E9 & % PM_;7

PM_{!C P}M

*PM_{= # P%=}

1 ;535@1D1>5<5;DB?>4989DE>745>71>=5>5>DE;1>:1B9:1B9?B29D&D5B<5298

418E<EI19DE

* &_PM_=PM_==1;1

.

!C

& %* _{;7 =}

. P_{= C}

2 >5B79;9>5D9;5<5;DB?>_#4981C9<;1>=5<1<E9@5BC1=11>

_{;7 = C}

#

% .

_#PM_!

3 >5B79@?D5>C91<5<5;DB?>₍4981C9<;1>=5<1<E9@5BC1=11>

%= _! =

(

#

*

4 >5B79D?D1<5<5;DB?>4981C9<;1>=5<1<E9@5BC1=11>

_#(

PM_!MPM_!

MPM_!

Contoh

8.2

Kata Kunci

• model atom Demokritus • model atom Dalton • Hukum Perbandingan Tetap • Hukum Lavoisier

• model atom Thomson • tetes minyak Millikan • model atom Rutherford • model atom Bohr • jari-jari Bohr • orbit stasioner

• energi elektron terkuantisasi

Tantangan

untuk Anda

Jika sebuah atom terdiri atas beberapa atom, apakah ada gaya elektrostatik antarelektron tersebut? Kemudian, bagaimana lintasan orbit elektron tersebut?

!9;1:1B9:1B95<5;DB?>I1>725B75B1;@141<9>D1C1>CD1C9?>5B*L25B1@1;18

;5<1:E1>5<5;DB?>D5BC52ED

9;5D18E9=1CC141>=E1D1>>I1%PM_;741>PM

!1G12

"5<1:E1>5<5;DB?>.4981C9<;1>45>71>=5>77E>1;1>@5BC1=11>

#

# % . #

* *

%=

= C ;7 =

. #

%*

!149;5<1:E1>5<5;DB?>D5BC52ED141<18P_{= C}

Contoh

8.3

)52E185<5;DB?>@1411D?=894B?75>25B@9>41841B95>5B795;C9D1C9&;5&

5B1@1;186B5;E5>C9@1>31B1>6?D?>>I1

(=()

9;5D18E9 5,

&

&

5, PM_{! &}

PM_!C

B. Atom Berelektron Banyak

1. Kelemahan Teori Atom Bohr

"5B1>7;1 41C1B D5?B9 1D?= ?8B C525>1B>I1 =5>713E ;5@141 =?45< 1D?= (ED85B6?B4

525B1@1 ;5<5=181> D5?B9 1D?= ?8B I1>7 41@1D 49C5=@EB>1;1> C589>77141@1D=5>I9>7;1@=9CD5B9D5>D1>71D?=141<18C52171925B9;ED *5?B9 1D?= ?8B D941; 41@1D =5>:5<1C;1> C531B1 B9>39 D5>D1>7 71B9C C@5;DBE= *5BED1=1 @5B25411> @1>:1>7 75<?=21>7 41> 9>D5>C9D1C 71B9C C@5;DBE= D5BC52ED

!E=<18 5<5;DB?> I1>7 D5B<921D 41<1= 1D?= 81>I1 C1DE 5<5;DB?> C589>771 E>DE; 1D?= 25B5<5;DB?> 21>I1; D941; 41@1D 497E>1;1> <179 ?>D?8>I1 1D?= 894B?75> =5=9<9;9 C1DE 5<5;DB?> 41> 71I1 5<5;DB?CD1D9C I1>7 141 81>I1 71I1 D1B9;=5>1B9; 1>D1B1 9>D9 1D?= 41> 5<5;DB?> 41@E> E>DE; 1D?= I1>7 =5=9<9;9 <5298 41B9 C1DE 5<5;DB?> C5<19> 71I1 D1B9;=5>1B9; =1C98 141 71I1 D?<1; 1>D1B1 5<5;DB?> I1>7 C1DE 41> 5<5;DB?> I1>7 <19>6/8=5>71219;1> 71I1

D?<1; 9>9

D?=1D?= 25B712E>7 =5=25>DE; J1D 1D1E =1D5B9 >D5B1;C9 1>D1B 1D?= =5>I5212;1> C961DC961D I1>7 25B2541 1>D1B1 E>CEB I1>7 C1DE 41>E>CEBI1>7<19>'5>I5212@5B25411>C961DC961DD5BC52EDD5B>I1D1 D941; =1=@E 49D5B1>7;1> ?<58 D5?B9 1D?= ?8B

B5;E5>C9@1>31B1>6?D?>4981C9<;1>45>71>=5>789DE>741>D5B<5298418E<E

_{5, 5, 5, !}

&

_{5, 5, 5, !}

&

1

! !

!C

J

$5>71@1 $/6496529C1=5>I9=@E<;1>218G1

5<5;DB?>=5BE@1;1>21791>41B91D?=

!5<1C;1> ;5<5=181>;5<5=181> 41<1= D5?B9 1D?= (ED85B6?B4

9DE>7 @1>:1>7 75<?=21>7 6?D?> D5B@5>45; 41> D5B@1>:1>741B945B5D1<=5B1D?=894B?75>

9DE>7@1>:1>775<?=21>76?D?>I1>749@1>31B;1> 1;921D DB1>C9C9 41B9& ;5& DB1>C9C9 ;5

=5BE@1;1>45B5D'1C385>

1B9 29<1>71> ;E1>DE= ;5 25B1@1;18 DB1>C9C9 I1>7 =5>781C9<;1>@1>:1>775<?=21>7L41B945B5D #I=1>

1B929<1>71>;E1>DE=;525B1@1;18DB1>C9C9I1>7=5>7 81C9<;1>@1>:1>775<?=21>7L41B945B5D1<=5B

9DE>7;535@1D1>5>5B79;9>5D9;5>5B79@?D5>C91<41> 5>5B79D?D1<5<5;DB?>@141?B29D;5?8BE>DE;1D?= 894B?75>

*5>DE;1><18@1>:1>775<?=21>76?D?>I1>749C5B1@ 1D?=894B?75>:9;15<5;DB?>25B5;C9D1C941B9&;5

D9>7;1D&

<5;DB?>1D?=894B?75>41@1D25B5;C9D1C9;5D9>7;1D 5>5B79I1>7<5298D9>7791;921D@5>5=21;1>*5>DE;1> 6B5;E5>C96?D?>3181I1I1>749C5B1@?<581D?=894B?75> 171B5<5;DB?>25B5;C9D1C941B9&;5D9>7;1D&

D?= 894B?75> 49D5=21; 45>71> 5<5;DB?> I1>7 =5=9<9;95>5B795,*5>DE;1>@1>:1>775<?=21>7 B1491C9I1>749@1>31B;1>71C894B?75>

Tes Kompetensi

Subbab

A

$?45<1D?=?8B=5<1>771B@B9>C9@;5D941;@1CD91>I1>749;5=E;1;1> ?<58,09,5),8.

1<1= =5>71D1C9 ;5;EB1>71> D5BC52ED 2525B1@1 69C9;1G1> C5@5BD9

6;09 +, 86.30,&63-.(5. !(;308=05 #*/86+05.,8 41>&(85,8

,09,5),8. =5=@5B;5>1<;1> =5;1>9;1 75<?=21>7 1D1E =5;1>9;1

;E1>DE= E>DE; =5>771=21B;1> ;51411> 5>5B79 5<5;DB?> 41<1= 1D?=

2. Bilangan Kuantum

<5;DB?>5<5;DB?> 25B75B1; =5>75<9<9>79 9>D9 41> 49;5<?=@?;;1> ;5 41<1= ;E<9D;E<9D $5>EBED D5?B9 1D?= ?8B E>DE; =5>I1D1;1> @?C9C9 5<5;DB?>81>I149@5B<E;1>C1DE!$&&#-&,-%I19DE29<1>71>;E1>DE=

ED1=1 5B2541 45>71> D5?B9 1D?= =5;1>9;1 ;E1>DE= @141 D5?B9 9>9 5<5;DB?>41<1=CE1DE1D?=4971=21B;1>41<1=5=@1D29<1>71>;E1>DE= I19DE

29<1>71> ;E1>DE= ED1=1 C9=2?<&

29<1>71> ;E1>DE= ?B29D1< C9=2?< 29<1>71> ;E1>DE= =17>5D9; C9=2?< %

29<1>71> ;E1>DE= C@9> C9=2?<%_C a. Bilangan Kuantum Utama (n₎

9<1>71> ;E1>DE= ED1=1 & =5>I1D1;1> 25C1B 5>5B79 D?D1< 41<1=

;E<9D 1D?= 1< 9>9 25B1BD9 E>DE; =5>75<E1B;1> 5<5;DB?> 41B9 ?B29D>I1 49@5B<E;1> C5:E=<18 5>5B79 >5B79 D5B5>418 49=E<19 41B9 ;E<9D " ;5=E491> ;E<9D # $ % & 41> C5D5BEC>I1 D?= 894B?75> =5=9<9;9 5>5B79 D?D1< C5CE19 45>71> @5BC1=11>

5, &

&

_M

+>DE; 9?> 5 _{I1>7 81>I1 =5=9<9;9 C52E18 5<5;DB?> 45>71>0}

25C1B 5>5B79 D?D1< 5<5;DB?>>I1 141<18

_{5, 5,}

&

0

& &

M

>5B79 D?D1< 5<5;DB?> 41<1= 1D?= 141<18 ;?>CD1> 41>,*#-&,!++!

?<58 29<1>71> ;E1>DE=& 9<1>71> ;E1>DE= ED1=1 =5=9<9;9 >9<19

C541>7;1> >1=1 ;E<9D C5CE19 45>71> 25C1B >9<19& I19DE

E>DE;&>1=1;E<9D"

E>DE;&>1=1;E<9D#

E>DE;&>1=1;E<9D$

E>DE;&>1=1;E<9D%41>C5D5BEC>I1

1>I1;>I1 5<5;DB?> =1;C9=1< I1>7 41@1D =5>79C9 C5D91@ ;E<9D 41@1D 49DE<9C;1> 45>71> BE=EC

&_{45>71>& M}

$9C1<>I1E>DE;;E<9D#29<1>71>;E1>DE=ED1=1>I1&!14921>I1;>I1

Gambar 8.12

Bentuk-bentuk orbital (a) Orbital s merupakan sebuah bola; (b) Orbital p memiliki bentuk seperti balon yang terpilin dua; (c) Orbital d memiliki bentuk seperti empat buah balon yang terpilin.

D?==5=9<9;95<5;DB?>*5>DE;1>14125B1@1;E<9D41>25B1@15<5;DB?>@141

C5D91@;E<9D

(=()

5B41C1B;1>;?>697EB1C95<5;DB?>BE=EC&_{4941@1D;1>;E<9D"=5=9<9;95<5;DB?>}

;E<9D#=5=9<9;95<5;DB?>41>5<5;DB?>=5>5=@1D9;E<9DC589>771'

!149@1411D?='D5B41@1D;E<9D"M#M$

Contoh

8.5

b. Bilangan Kuantum Orbital (₎

>41D5<18=5>75D18E9218G15<5;DB?>25B?D1C9=5>79D1B99>D91D?= @141 ?B29D>I1 9<1>71> ;E1>DE= ?B29D1< =5>I1D1;1> =?=5>DE= CE4ED 5<5;DB?> I1>7 25B?D1C9 D5B8141@ @?B?C 9>D9 1D?= $5<1<E9 @5BC1=11>

*'!&* 25C1B =?=5>DE= CE4ED 5<5;DB?> 141<18

M

45>71>

PM!C

%9<19 25B71>DE>7 @141 >9<19 29<1>71> ;E1>DE= ED1=1_& I19DE =E<19 41B9 >?< C1=@19 45>71> & M 1D1E 49>I1D1;1> C521719

_&M

)5D91@ ?B29D1< =5=9<9;9 >1=1 41> 25>DE; D5BD5>DE %1=1 ?B29D1< I1>7 49>I1D1;1> ?<58 >9<19 141<18

49>1=1;1> CE2;E<9D_{++ *(} 49>1=1;1> CE2;E<9D_((*!&!($ 49>1=1;1>CE2;E<9D_!-+ 49>1=1;1> CE2;E<9D -&%&,$

(4)(8 =5=@5B<981D;1> 25>DE; C5D91@ ?B29D1< 9<1>71> ;E1>DE=

?B29D1< :E71 C5B9>7 49C52ED 45>71> 29<1>71> ;E1>DE= 1J9=ED

3 2

1

*5>DE;1>=?=5>DE=CE4ED5<5;DB?>E>DE; PM_!C

(=()

5>71>=5>77E>1;1>!,89(4((5@=1;1

!C

PM_!C

Contoh

8.6

Atom adalah partikel yang sangat

kecil sehingga tidak dapat diamati dengan cahaya tampak.

Menggambarkan struktur atom merupakan pekerjaan yang sangat menarik. Untuk mempelajari atom, para ahli mempelajarinya dari cahaya yang diesensikan oleh atom-atom dari unsur murni. Spektrum cahaya tersebut mengandung informasi tentang struktur, atom, energi ionisasi, dan lain-lain.

Atom are tiny, and not observable using visible light. Trying to figure out the structure of atoms is a facinating task. A major source of information about atoms comes from a study of the light emitted by the atoms of a pure material. The spectrum of the light give information about atom structure, ionization energy, etc.

Informasi

untuk Anda

$(),3

$?=5>DE=)E4ED&B29D1<<5;DB?> %1=1

+ ( !

CE2;E<9D

"51411>=?=5>DE=CE4ED?B29D5<5;DB?>49>1=1;1>45>71>CE2;E<9D I19DE CE2;E<9D+ E>DE; ;51411> CE2;E<9D₍ E>DE; ;51411> CE2;E<9DE>DE;;51411>41>C5D5BEC>I1'5B81D9;1>_$(),3

5C1B>I1=?=5>DE=CE4ED?B29D1<5<5;DB?>E>DE;45B5D@5BD1=1I1>7 25B1C1< 41B9 C@5;DBE= I19DE 45B5D + + *(( (*!&!($ !-+ 41> -&%&,$ 141<18

)E2;E<9D+25C1B=?=5>DE=CE4ED>I1

)E2;E<9D(25C1B=?=5>DE=CE4ED>I1

)E2;E<9D25C1B=?=5>DE=CE4ED>I1

)E2;E<9D25C1B=?=5>DE=CE4ED>I1

12E>71> 29<1>71> ;E1>DE= ED1=1 & 45>71> 29<1>71> ;E1>DE=

?B29D1< 291C1>I1 497E>1;1> E>DE; =5>I1D1;1> ;51411> 5<5;DB?> 41<1= CE1DE 1D?= $9C1<>I1 + =5>I1D1;1> ;51411> 5<5;DB?> 45>71> &C541>7;1>=5>I1D1;1>;51411>5<5;DB?>45>71>_& 41>45=9;91>C5D5BEC>I1

1>I1;>I1 5<5;DB?> =1;C9=E= 41<1= CE2;E<9D+ 141<18 5<5;DB?>

41<1=CE2;E<9D(141<185<5;DB?>CE2;E<9D141<185<5;DB?>CE2;E<9D 141<185<5;DB?>41>C5D5BEC>I1C5@5BD9$(),3

c. Bilangan Kuantum Magnetik (m

)

&<58 ;1B5>1 5<5;DB?> 25B=E1D1> <9CDB9; 75B1; 5<5;DB?> 41@1D 49@5>71BE89 ?<58 =541> =17>5D9; ;921D>I1 5<5;DB?> I1>7 25B75B1; 49 41<1= ?B29D>I1 =5=9<9;9 =?=5> =17>5D9; $?=5> =17>5D9; 41B9 5<5;DB?> 9>9 49>I1D1;1> 45>71> 29<1>71> ;E1>DE= =17>5D9; =

9<1>71> ;E1>DE= =17>5D9; =5>I1D1;1> 1B18 =?=5>DE= CE4ED 5<5;DB?>!9;149D5D1@;1>1B18=541>=17>5D9;<E1BC5:1:1B45>71>CE=2E

0 ;5=E>7;9>1> 41<1= 1B180 49>I1D1;1> 45>71> @5BC1=11>

0

%

M

9<1>71> ;E1>DE= =17>5D9; = =5=9<9;9 >9<19 1>D1B1 M41>

$9C1<>I1

E>DE;81B71=1D1ED5B41@1D?B29D1< E>DE;81B71=

M41>1D1ED5B41@1D?B29D1<

E>DE;81B71=

MM1D1ED5B41@1D?B29D1<

5C1B_J41@1D 49<981D @141(4)(8

$(),3 1>I1;<5;DB?>@141

)5D91@)E2;E<9D

)E2;E<9D + (

1>I1;%₅

1>I1;5<5;DB?>

P21>I1;=

2

0

–2

–

m= 2

m= 1

m= 0

m= 1

m= 2

= 2

L_z

Gambar 8.13

Arah dan besar momentum sudut

'141 (4)(8 49@5B<981D;1> F5;D?BF5;D?B =?=5>DE= CE4ED

45>71> 29<1>71> ;E1>DE= ?B29D1< D5BD5>DE 9<1>71> ;E1>DE= =17>5D9; :E71 =5>771=21B;1> ;535>45BE>71> ;54E4E;1> ?B29D1< 49 41<1= BE1>7 C5@5BD9 D5B<981D @141(4)(8

3 2

1

Gambar 8.14

Arah momentum sudut-sudut orbital s, p, dan d

(a) orbital hanya memiliki 1 kemungkinan arah; (b) orbital p memiliki 3 kemungkinan; dan (c) orbital d memiliki 5 kemungkinan.

z z z

_Z 0

0

_Z

2

h

Z ₂

h

1 2 Z

h

_Z

2

h

_Z 0 _Z0

_Z

2

h

_Z h

Gambar 8.15

Elektron dalam subkulit s, p, dan d.

3

2 1

Gambar 8.16

Arah perputaran dan vektor momentum sudut S.

d_xy d_xz d_yz d_x2_y2d_x2_y2

s p_x p_y p_z

1

2

1 2

Z ₄h

S p

Z ₄h

S p

S

$(),3

"5C9=@E<1>"55=@1D9<1>71>"E1>DE=

9<1>71>;E1>DE=ED1=1 &

9<1>71>;E1>DE=?B29D1< >M

9<1>71>;E1>DE==17>5D9; % MMM

9<1>71>;E1>DE=C@9> %_C

03(0?(5.07,82,5(52(5

(4( 6:(90

*E>:E;;1><18218G1;E<9D=1;C9=E=49D5=@1D9?<585<5;DB?> (=()

"E<9D=5=9<9;929<1>71>;E1>DE=&

+>DE;&>9<1929<1>71>;E1>DE=?B29D1<>I141>

+>DE;D5B41@1DP2E18>9<19_%

+>DE;D5B41@1DP2E18>9<19_%

!E=<18 2E18>9<19%

Contoh

8.7

d. Bilangan Kuantum Spin (m s)

9<1>71> ;E1>DE= C@9> %_C =5>I1D1;1> 1B18 @5B@ED1B1> 5<5;DB?>

D5B8141@ CE=2E>I1 C@9> I1>7 41@1D =5>9=2E<;1> =?=5> =17>5D9; %9<1929<1>71>;E1>DE=C@9>1414E1I19DE%_C E>DE;1B18@ED1B1> 5<5;DB?> ;5 ;1>1> 41>%_C E>DE; 1B18 @ED1B1> 5<5;DB?> ;5 ;9B9

$1C9>7=1C9>7 49DE<9C 45>71> >?D1C9 1B18 @1>18 ;5 1D1C 41> ;5 21G18 )E2;E<9D + =5=9<9;9 C52E18 ?B29D1< 41> @1<9>7 21>I1; 499C9 4E1 5<5;DB?>

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(:(:(5

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@141(4)(8 !9;1 C52E18 5<5;DB?> 25B=1CC1% 25B75B1; 41<1=

?B29DI1>725B:1B9:1B9*45>71>6B5;E5>C9=?=5>DE=CE4EDI1>749=9<9;9

5<5;DB?> 141<18 %₅.* %₅* _{C541>7;1> 1B18 =?=5>DE=>I1}

25BCE4ED D5B8141@CE=2E05C1B>I11BECI1>7D9=2E<1;921D75B1;1>

5<5;DB?> 494569>9C9;1> C521719 21>I1;>I1 =E1D1> I1>7 =5>71<9B D91@ 45D9; !149 1BEC I1>7 49D9=2E<;1> 75B1;1> 5<5;DB?> ) )

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25BE218 5C1B>I1 D?BC9 1;921D =541> =17>5D 141<18

P M

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%

C9> M

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5C1B 5>5B79 =17>5D =5=5>E89 @5BC1=11>

3?C 3?C

%

Gambar 8.17

Gerak elektron dalam medan magnet B.

Tantangan

untuk Anda

Tentukan besar momentum sudut untuk subkulit:

a. f; b. g; c h.

L Z

Gambar 8.18

Arah momentum sudut elektron.

L_z = m_e

Z

Gambar 8.19

Terpecahnya garis spektrum akibat medan magnet.

5C1B1> % 49>1=1;1> @5=21>49>7 79B?=17>5D9; /*'%&,! *,!'

'5B81D9;1>(4)(8 1=21B D5BC52ED =5>E>:E;;1> 218G1

3?C %

C589>771;9D1@5B?<58

%

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%

%

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%

M

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%

%

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%

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%

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M

& &

M

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P !C P = C

5, P ! 5,

P₌

'5BE2181>5>5B79 PM_{! **P}M_!

!14925B41C1B;1>!,89(4((5@49@5B?<58

P = P !C P = C

PM!PM=

3. Sifat Atom

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'5>79C91> 5<5;DB?> 49=E<19 41B9 D9>7;1D 5>5B79 D5B5>418 "?>697EB1C9 I1>7=1>D1@D5B41@1D@141CE2;E<9DI1>7D5B9C9@5>E8!9;1CE2;E<9DD5<18 D5B9C9 @5>E8 C9C1 5<5;DB?> 1;1> =5>79C9 CE2;E<9D C5<1>:ED>I1 $9C1<>I1 141 C1DE 5<5;DB?> I1>7 D5B9C9 @141 CE2;E<9D D5B<E1B =1;1 5<5;DB?> 9>9 35>45BE>7 =E418 <5@1C CE@1I1 1D?= =5>:149 CD129< '5<5@1C1> 5<5;DB?> 9>9 41@1D D5B:149 45>71> 31B1 25B9;1D1> 45>71> 1D?= <19>

)521719 3?>D?8 1D?= %1 I1>7 =5=9<9;9 ;?>697EB1C9

+ (

+ +

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I19DE@141?B29D++>DE;=5>:149CD129<1D?=%135>45BE>7=5<5@1C;1>

C1DE 5<5;DB?> D5B<E1B>I1 I1>7 D5B41@1D @141 ;E<9D )521<9;>I1 :9;1

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;E<9D D5B9C9 5<5;DB?> "E<9D D5B9C9 =1;C9=1< 5<5;DB?> 25B1BD9

D5B41@1D;5;EB1>71>5<5;DB?>C589>7711D?=$>35>45BE>7=5>5B9=1 C1DE 5<5;DB?> 41B9 1D?= <19> CE@1I1 =5>:149 CD129< D?= %1 41> $> 41@1D 25B712E>7 =5>:149 =?<5;E< I1>7 CD129< ;1B5>1 @141 C5D91@ 1D?= 1;1> =5=9<9;9 ;E<9D 1D1E CE2;E<9D I1>7 D5B9C9 @5>E8 ?<58 5<5;DB?>

Contoh

8.8

Gambar 8.21

(a) Atom Mn menerima elektron dari atom Na.

(b) Atom Na yang tidak stabil.

2 1

elektron dilepas

Na Na

Gambar 8.20

(a) Atom Na, kulit terluarnya terisi satu elektron.

(b) Atom Na yang stabil setelah melepas elektron terluarnya.

2 1

e

Na Mn

)961DC961D;9=91E>CEB1<;1<9141<1881>I1=5=9<9;9C1DE5<5;DB?>49 ;E<9D D5B<E1B 1@1D<18 49D5B;1 218G1 E>CEBE>CEB 9>9 =5=9<9;9 5>5B79 9?>9C1C9I1>7B5>418C1>71D=E418=5<5@1C;1>5<5;DB?>)5=1;9>25C1B :1B9:1B91D?=5>5B799?>9C1C9>I1C5=1;9>B5>4185>71>;1D1<19>E>CEB E>CEB 1<;1<9 9>9 C1>71D B51;D96 1D1E =E418 C5;1<9 =5=25>DE; C5>I1G1 +>CEBE>CEB 9>9 =E418 C5;1<9 25B51;C9 45>71> ?;C975> 49 E41B1 DE<18 C5212>I1 <?71=<?71= 1<;1<9 81BEC 49C9=@1> 41<1= =9>I1; D1>18 41> D941; 2?<58 49291B;1> <1=1 49 E41B1 D5B2E;1

b. Afinitas Elektron

!&!,+$#,*'&$ &*!/&!$(+#&"!#+-,-,'%%&&#( +- $#,*'&D?=1D?=I1>7=5=9<9;971I1D1B9;=5>1B9;1>D1B9>D9>I1

;539< =5>E>:E;;1> 218G1 169>9D1C 5<5;DB?>>I1 :E71 ;539< 1<81< I1>7 =5=5>71BE89 25C1B ;539<>I1 169>9D1C 5<5;DB?> 141<18 :E=<18 =E1D1> 41<1= 9>D9 :1B1; 1>D1B 9>D9 41> :E=<18 5<5;DB?> 41<1= 1D?=

1<1= C1DE 7?<?>71> 41B9 1D1C ;5 21G18 81B71 169>9D1C 5<5;DB?> C5=1;9>;539<1<1=C1DE@5B9?4541B9;9B9;5;1>1>81B71169>9D1C5<5;DB?> C5=1;9>25C1BD?=1D?=I1>7C5=1;9>=E418=5>1>7;1@5<5;DB?>1;1> =5=9<9;981B71169>9D1C5<5;DB?>I1>7C5=1;9>25C1B+>DE;=5>771=21B;1> D9>7;1D ;5=E4181> CE1DE 1D?= =5>1>7;1@ 5<5;DB?> 497E>1;1> 25C1B1> 5<5;DB?>571D9F9D1C 1D1E ;55<5;DB?>571D961> )E1DE E>CEB I1>7 =5=9<9;9 5>5B799?>9C1C9D9>779=5=9<9;95<5;DB?>571D9F9D1CD9>779BD9>I1=5=9<9;9 ;535>45BE>71> 25C1B E>DE; =5>1>7;1@ 5<5;DB?>

+>CEBI1>7=E418=5<5@1C;1>5<5;DB?>41>=5=25>DE;9?>@?C9D96 49>1=1;1> E>CEB I1>7 5<5;DB?@?C9D96 )521<9;>I1 E>CEB I1>7 =E418 =5>1>7;1@ 5<5;DB?> 41> =5=25>DE; 9?> >571D96 49>1=1;1> E>CEB 5<5;DB?>571D96'141E=E=>I1E>CEBE>CEBI1>7=5=9<9;95<5;DB?>F1<5>C9 ;EB1>7 41B9 5=@1D 25BC961D 5<5;DB?@?C9D96 41@E> E>CEBE>CEB I1>7 =5=9<9;95<5;DB?>F1<5>C9<529841B95=@1D25BC961D5<5;DB?>571D9669>9D1C 2525B1@1 E>CEB D5B<981D @141$(),3

c. Elektron Valensi

<5;DB?> I1>7 @1<9>7 25B@5B1> 41<1= =5>5>DE;1> C961D 9C9;1 41> "9=91 141<18 5<5;B?> I1>7 25B141 @141 ;E<9D @1<9>7 <E1B <5;DB?> 9>9 291C1 49C52ED$#,*'& .$&+! ;1> D5D1@9 E>DE; ;E<9D I1>7 @5>E8

5<5;DB?>>I1 D941; DEBED C5BD1 41<1= =5>5>DE;1> C961DC961D D5BC52ED 1B9CECE>1>25B;1<1E>CEBE>CEB45>71>C9CD5=@5B9?49;41<1=C1DE 7?<?>71> 1D?=1D?= D5BC52ED =5=9<9;9 5<5;DB?> F1<5>C9 I1>7 C1=1 $9C1<>I11D?=#9I1>7=5=9<9;95<5;DB?>C1<18C1DE>I1141<185<5;DB?> F1<5>C9 5=9;91> 81<>I1 1D?= " =5=9<9;9 5<5;DB?> C1<18 C1DE>I1 141<185<5;DB?>F1<5>C9"5C9=@E<1>>I11D?=#941>1D?="I1>725B141 41<1= C1DE 7?<?>71> =5=9<9;9 C1DE 5<5;DB?> F1<5>C9

M

#9 M

%1 M

M

M

)9 M

% M

' M

& M

M

< M

B M

M $(),3

69>9D1C525B1@1+>CEB

%59;8 -050:(9

3,2:8652463

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Mari Mencari Tahu

• teori atom Bohr • bilangan kuantum utama • bilangan kuantum orbital • bilangan kuantum magnetik • bilangan kuantum spin • prinsip Pauli

• aturan Aufbau • kaidah Hund • spektrum emisi • spektrum absorsi • garis-garis Praunhoffer • energi ionisasi • afinitas elektron • elektron valensi

Sumber:General Chemistry, 1990

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Tes Kompetensi

Subbab

B

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Rangkuman

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Peta Konsep

Setelah mempelajari bab ini, tentu Anda telah memahami bagaimana perkembangan model atom sampai terakhir digunakan model atom Bohr. Anda juga tentu telah paham bagaimana kedudukan dan sifat elektron dalam bab inti. Dari semua materi yang ada dalam bab ini, bagian manakah yang menurut Anda sulit dipahami? Coba Anda diskusikan bersama teman atau guru Fisika Anda.

Refleksi

Sekarang Anda menjadi tahu bahwa rekayasa atom untuk dimanfaatkan dalam kehidupan bergantung sekali pada elektron valensi atom. Coba Anda cari manfaat lain yang diperoleh setelah mempelajari bab ini, kemudian diskusikan konsep Fisika apa yang terkandung di dalamnya bersama teman Anda.

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