Question

It is desired that the radiation energy emitted by a light source reach a maximum at 470 nm. Determine the temperature of this light source and the fraction of radiation it emits in the visible range (i.e 0.40–0.76 µm)

Answer 1

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