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Anastasy [175]
3 years ago
12

It is desired that the radiation energy emitted by a light source reach a maximum at 470 nm. Determine the temperature of this l

ight source and the fraction of radiation it emits in the visible range (i.e 0.40–0.76 µm)
Physics
1 answer:
zalisa [80]3 years ago
4 0

Answer:

The temperature of this light is 6148 K

Explanation:

Given:

Maximum wavelength \lambda _{m}= 470 \times 10^{-9} m

According to the wien's displacement law,

Maximum wavelength of light is inversely proportional to the temperature of body.

    \lambda _{m } T = 2.89 \times 10^{-3}

Now find temperature,

    T = \frac{2.89 \times 10^{-3} }{470 \times 10 ^{-9} }

    T = 6148 K

Therefore, the temperature of this light is 6148 K

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