Analisis Data unit 5

A. Analisis Data d

=

_N¹

d

=

₆₀₀¹_mm

Kegiatan 1. Menggunakan rumus Balmer untuk menentukan konstanta Rydberg

1. Warna Biru

a. Panjang Gelombang λ=dsinθ

2 λ=dsin∆ θ

2 λ= 1

600sin 17,00 λ=¿0,0005 mm λ=¿ 500 nm

b. Ketidakpastian Panjang Gelombang λ=dsinθ_kiri+θ_kanan

2

∆λ =

|

^{∂ θ∂ λkiri}

|

^{∆ θkiri}

+ |

^{∂ θ∂ λkanan}

|

^{∆ θkanan}

∆λ =

|

^{∂sinθkiri+2∂ θθkanankiri (n . N)−1}

|

^{∆ θkiri+}

|

^{∂sinθkiri+∂ θ2θkanankanan(n . N)−1}

|

^{∆ θkanan}

∆λ =

|

^{12sinθkiri+θ2 kanan(n . N)−1}

|

^{∆ θkiri+}

|

^{12sinθkiri+θ2kanan(n . N)−1}

|

^{∆ θkanan}

λ∆ λ

λ

= |12sincosθθkirikiri+2+θ2θkanankanan((n . Nn . N)−1)−1|∆ θkiri+

|

^{12sincosθθkirikiri+2+θθ2kanankanan((n . Nn . N)−1)−1}

|

^{∆ θkanan}

∆ λ

λ =

|

^{12cossinθθkirikiri++22θθkanankanan}

|

^{(∆ θkiri + ∆ θkanan)}

∆λ =

|

^{2 tan∆ θkiriθ+kiri∆ θ+2θkanankanan}

|

^λ

∆λ =

|

^{2 tan∆ ∆ θ∆ θ2}

|

^λ

∆λ =

|

^{2 tan 170,0 085˚}

|

^{500 nm}

∆λ =

|

^120,027802

|

^{500 nm}

∆λ = |^{0,0 13901}| ^{500 nm}

∆λ = 6,950561814 nm

c. Kesalahan Relatif KR = ∆ λ

λ ×100% KR = 6,950561814nm

500nm ×100%

KR = 1,39 % d. Pelaporan Fisika

PF = |^{λ ± ∆ λ}|^nm PF = |^500±6,95|^nm 2. Warna Merah

a. Panjang Gelombang

λ=dsinθ 2 λ=dsin∆ θ

2 λ= 1

600sin 23,59 λ=¿0,0007 mm

λ=¿ 700 nm

b. Ketidakpastian Panjang Gelombang

∆λ =

|

^{2 tan∆ ∆ θ∆ θ2}

|

^λ

∆ λ=

|

^{2 tan∆ ∆ θ∆ θ2}

|

∆λ =

|

^{12 tan 230,0085,59}

|

^{700 nm}

∆λ =

|

^120,019465

|

^{700 nm}

∆λ = |^{0,0 09732}| ^{700 nm}

∆λ = 6,812746813 nm

c. Kesalahan Relatif KR = ∆ λ

λ ×100% KR = 6,812746813nm

700nm ×100%

KR = 0,97 % d. Pelaporan Fisika

PF = |^{λ ± ∆ λ}|^nm PF = |^700±6,81|nm 3. Konstanta Rydberg

a. Untuk warna 1 (Biru) 1

λ = R_H

[

^{(n1f)2−(n1i)2}

]

R_H= 1

λ

[

^{(n1f)2−(n1i)2}

]

R_H= 1

500nm 1 2²−1

5²

R_H= 1

500nm1 4− 1

25

R_H= 1

500nm(0,21) RH = 4,761904762 nm b. Untuk warna 2 (Merah)

1

λ = R_H

[

^{(n1f)2−(n1i)2}

]

R_H= 1

λ

[

^{(n1f)2−(n1i)2}

]

R_H= 1

700nm 1 2²−1

5²

R_H= 1

700nm1 4− 1

25

R_H= 1

700nm(0,21)

RH = 4,761904762 nm

4. Konstanta Rydberg Rata-Rata R=R₁+R₂

2

R=(4,761904762+4,761904762)nm 2

R=4,76190nm

%diff=

|

^{RRTeoriTeori−+R2RPraktikumPraktikum}

|

^×100%

%diff=

|

^{((1,0971,097××101077−4+4,761902,76190××1010−9−9))mm}

|

^×100%

%diff = 199,9 % 5. Konstanta Planck

R=2π²k²me⁴ ch³ h³=2π²k²me⁴

cR³

h=

√

^{3 2π2cRk2me4}

h =

√

^{3 2(3,14)2(9}

(

^{3××101098)2}

)

^{((4,761909,1×10×10−31)(1−9,)6×10−19)4}

h =

√

^{3 2(9,8596)(81×}

(

^3×1010188)(

)

^{(4,76190×9,1×10−3110)(−96,5536) ×10−76)}

h =

√

^{3 (}1.597,2552×10¹⁸)(9,1×10⁻³¹)(6,5536×10⁻⁷⁶)

(

3×10⁸

)

(4,76190×10⁻⁹) h =

√

³ 95.256,722276×10⁸⁹

(

^3×108

)

(4,76190×10⁻⁹) h =

√

³ 95.256,722276×10⁸⁹

14,2857×10⁻¹ h =

√

^{36.667,977×1090}

h = 2,222659 ×10⁹³ J.s

%diff =

|

^{hhTeoriTeori−h+h2PraktikumPraktikum}

|

^×100%

%diff =

|

⁽

(

^{66,,6262xx10103434−+2,2226592,2226592 ××10109393)}

)

^{J . sJ . s}

|

^×100%

%diff = 1 99,9 %

Dengan analisis yang sama maka diperoleh tabel 4.3 berikut Tabel 4.3 Hasil Analisis Data Kegiatan 1 pada orde Spektrum 1 Warna

Spektru m

Panjang Gelombang

(nm)

Konstanta Rydberg

%diff Konstanta

Rydberg

Konstanta Planck

%diff Konstanta

Planck

Biru 500 4,76190476

2 199,9 % 2,222659

×10⁹³

199,9 %

Merah 700 4,76190476

2 199,9 % 2,222659

×10⁹³

199,9 %

Kegiatan 2. Menentukan jenis unsur sumber cahaya yang tidak diketahui berdasarkan spektrum garis yang terbentuk

Lampu 1

1. Warna Ungu

a. Panjang Gelombang λ=dsinθ

2 λ=dsin∆ θ

2 λ= 1

600sin 16,2 λ=0,0005 mm λ=500 nm

b. Ketidakpastian Panjang Gelombang

∆λ =

|

^{2 tan∆ ∆ θ∆ θ2}

|

^λ

∆ λ=

|

^{2 tan∆ ∆ θ∆ θ2}

|

∆λ =

|

^{12 tan 160,0085,2˚}

|

^{500 nm}

∆λ =

|

^120,029267

|

^{500 nm}

∆λ = |^0,014633362| ^{500 nm}

∆λ = 7,31668115 nm

c. Ketidakpastian KR = ∆ λ

λ ×100%

KR = 7,31668115nm

500nm ×100%

KR = 1,46 % d. Pelaporan Fisika

PF = |^{λ ± ∆ λ}|^{×103 nm} PF = |^500±7,32| ^nm 2. Warna Hijau

a. Panjang Gelombang λ=dsinθ

2 λ=dsin∆ θ

2 λ= 1

600sin 18,66 λ=¿ 0,0005 mm λ=¿ 500 nm

b. Ketidakpastian Panjang Gelombang

∆λ =

|

^{2 tan∆ ∆ θ∆ θ2}

|

^λ

∆ λ=

|

^{2 tan∆ ∆ θ∆ θ2}

|

∆λ =

|

^{12 tan 180,0085,66˚}

|

^{500 nm}

∆λ =

|

^120,02517

|

^{500 nm}

∆λ = |0,012585008| ^{500 nm}

∆λ = 6,29250414nm

c. Ketidakpastian KR = ∆ λ

λ ×100%

KR = 6,29250414nm

500nm ×100% KR = 1,26 %

d. Pelaporan Fisika PF = |^{λ ± ∆ λ}| ^nm PF = |^500±1,25| ^nm Lampu 2

1. Warna Neon

a. Panjang Gelombang λ=dsinθ

2 λ=dsin∆ θ

2 λ= 1

600sin 15,485 λ=¿ 0,0004 mm λ=¿ 400 nm

b. Ketidakpastian Panjang Gelombang

∆λ =

|

^{2 tan∆ ∆ θ∆ θ2}

|

^λ

∆ λ=

|

^{2 tan∆ ∆ θ∆ θ2}

|

∆λ =

|

^{12 tan 15,4850,0085 ˚}

|

^{400 nm}

∆λ =

|

^120,030681

|

^{400 nm}

∆λ = |^{0,0 15340599}| ^nm

∆λ = 6,13623975 nm

c. Ketidakpastian KR = ∆ λ

λ ×100%

KR = 6,13623975nm

400nm ×100%

KR = 1,53 % (3 AP) d. Pelaporan Fisika

PF = |^{λ ± ∆ λ}|^nm PF = |^400±6,14| nm 2. Warna Hijau

e. Panjang Gelombang λ=dsinθ

2 λ=dsin∆ θ

2 λ= 1

600sin 20,005 λ=¿ 0,0006 mm λ=600 nm

f. Ketidakpastian Panjang Gelombang

∆λ =

|

^{2 tan∆ ∆ θ∆ θ2}

|

^λ

∆ λ=

|

^{2 tan∆ ∆ θ∆ θ2}

|

∆λ =

|

^{12 tan 20,0050,0085 ˚}

|

^{600 nm}

∆λ =

|

^120,023347

|

^{600 nm}

∆λ = |^0,011673609| ^{600 nm}

∆λ = 7,00416555 nm g. Ketidakpastian

KR = ∆ λ

λ ×100%

KR = 7,00416555nm

600nm ×100% KR = 1,17 %

h. Pelaporan Fisika PF = |^{λ ± ∆ λ}| ^nm PF = |^600±7,00|^nm

Tabel 4.4 Analisis Panjang Gelombang Kegiatan 2 Beberapa Jenis Atom

Lamp u ke-

Warna Spektru

m

λ (nm) ∆λ (nm) KR (%) PF (nm)

1 Ungu 500 7,31668115 1,46 │500 ± 7,32│

Hijau 500 6,29250414 1,26 │500 ± 6,29│

2 Neon 400 6,13623975 1,53 │400 ± 6,14│

Hijau 600 7,00416555 1,17 │600 ± 7,00│