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Makovka662 [10]
3 years ago
9

The element in a fluorescent lightbulb that absorbs UV light and releases visible light energy is ____?

Physics
2 answers:
Agata [3.3K]3 years ago
8 0
You're answer will be a phosphor 
The element in a florescent lightbulb that absorbs UV light and releases visible light energy is --a phosphor--. 
Mnenie [13.5K]3 years ago
3 0
So in a fluorescent light bulb the mercury atoms release the UV light. This UV light is absorbed by the phosphorous powder coating inside and this releases the visible light.
So your answer is phosphorous powder.
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In a physics lab, light with a wavelength of 490 nm travels in air from a laser to a photocell in a time of 17.5 ns . When a sla
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Answer:

196 nm

Explanation:

Given that

Value of wavelength, = 490 nm

Time spent in air, t(a) = 17.5 ns

Thickness of glass, th = 0.8 m

Time spent in glass, t(g) = 21.5 ns

Speed of light in a vacuum, c = 3*10^8 m/s

To start with, we find the difference between the two time spent

Time spent on glass - Time spent in air

21.5 - 17.5 = 4 ns

0.8/(c/n) - 0.8/c = 4 ns

Note, light travels with c/n speed in media that has index of refraction

(n - 1) * 0.8/c = 4 ns

n - 1 = (4 ns * c) / 0.8

n - 1 = (4*10^-9 * 3*10^8) / 0.8

n - 1 = 1.2/0.8

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n = 1.5 + 1

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490 / 2.5 = 196 nm

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