Firefly bioluminescence has a peak wavelength of around 600nm. From Wien’s blackbody radiation displacement law calculate the temperature if this were a blackbody spectrum. Could this be a blackbody spectrum and why or why not?

brasocas6
2022-08-13
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Joe Sheppard

Answered 2022-08-14
Author has **12** answers

Given:

The peak wavelength $({\lambda}_{peak})=600\text{}nm$

According to Wien’s blackbody radiation displacement law,

${\lambda}_{peak}T=2.898\times {10}^{-3}\text{}mK$

The peak possible wavelength of blackbody is 579.6 nm that occurs at 5000 K temperature,

${\lambda}_{peak}T=2.898\times {10}^{-3}\text{}mK\phantom{\rule{0ex}{0ex}}{\lambda}_{peak}=\frac{2.898\times {10}^{-3}}{5000}\phantom{\rule{0ex}{0ex}}{\lambda}_{peak}=579.6\text{}nm$

Now the temperature for given wavelength,

${\lambda}_{peak}T=2.898\times {10}^{-3}\text{}mK\phantom{\rule{0ex}{0ex}}T=\frac{2.898\times {10}^{-3}}{600\times {10}^{-9}}\phantom{\rule{0ex}{0ex}}T=4830\text{}nm$

Answer:

The given wavelength is black body spectrum having peak wavelength of 600 nm and temperature of 4830 K.

The peak wavelength $({\lambda}_{peak})=600\text{}nm$

According to Wien’s blackbody radiation displacement law,

${\lambda}_{peak}T=2.898\times {10}^{-3}\text{}mK$

The peak possible wavelength of blackbody is 579.6 nm that occurs at 5000 K temperature,

${\lambda}_{peak}T=2.898\times {10}^{-3}\text{}mK\phantom{\rule{0ex}{0ex}}{\lambda}_{peak}=\frac{2.898\times {10}^{-3}}{5000}\phantom{\rule{0ex}{0ex}}{\lambda}_{peak}=579.6\text{}nm$

Now the temperature for given wavelength,

${\lambda}_{peak}T=2.898\times {10}^{-3}\text{}mK\phantom{\rule{0ex}{0ex}}T=\frac{2.898\times {10}^{-3}}{600\times {10}^{-9}}\phantom{\rule{0ex}{0ex}}T=4830\text{}nm$

Answer:

The given wavelength is black body spectrum having peak wavelength of 600 nm and temperature of 4830 K.

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