A quantum system has three energy levels, so three wavelengths appear in its emission spectrum. the shortest observed wavelength
1 answer:
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Answer:
Y = 3.87 x 10⁻³ m = 3.87 mm
Explanation:
This problem can be solved by using Young's double-slit experiment formula:
where,
Y = fringe spacing = ?
L = slit to screen distance = 2 m
λ = wavelength of light = 580 nm = 5.8 x 10⁻⁷ m
d = slit width = 0.3 mm = 3 x 10⁻⁴ m
Therefore,
<u>Y = 3.87 x 10⁻³ m = 3.87 mm</u>
Answer:
775.48 W
Explanation:
given,
diameter of disk = 0.6 cm
length of the disk = 0.4 m
T₁ = 450 K T₂ = 450 K T₃ = 300 K
= 1.33
now,
the value of view factor (F₁₂)corresponding to 1.33
F₁₂ = 0.265
F₁₃ = 1 - 0.265 = 0.735
now,
net rate of radiation heat transfer from the disk to the environment:
= 2 F₁₃ A₁ σ (T₁⁴ - T₃⁴)
= 2 x 0.735 x π x (0.3)² x (5.67 x 10⁻⁸ W/m²) (450⁴ - 300⁴)
= 775.48 W
Net radiation heat transfer from the disks to the environment = 775.48 W