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Pembahasan Fisika Kuantum
1. Diket: kalor radiasiE₁ = 16 E₀
Dit: T₁Jwb:
E = Q/t = σ A T⁴Q/t ≈ T⁴
E₁/E₀ = T₁⁴/T₀⁴16 E₀/E₀ = T₁⁴/T₀⁴T₁⁴/T₀⁴ = 16T₁⁴ = 16 T₀⁴→ diakar pangkat 4
T₁ = ⁴√(16 T₀⁴)T₁ = 2 T₀
2. T320T₁ = 47 ºC = 320 KT₂ = 367 ºC = 640 K
intensitas I = E/(t A) = e σ T⁴
I₁....e σ T₁⁴-- = ----------- I₂....e σ T₂⁴
...= (T₁/T₂)⁴
...= (320 / 640) ⁴
...= (1/2)⁴
...= 1 : 16..<-- jawaban
3. Ss
4. Ss
5. SS
6. SS
7. Hubungan panjang gelombang maksimum dan suhu :λmaks . T = konstanta Wien(6000 x 10− 10 ) . T = 2,90 x 10− 3
T = 4.833 Kelvin