Asked by anna
HW Help!!!!
An object at 4500 K emits the blackbody spectrum shown here.
This object emits the most light in the __?__ part of the visible spectrum.
An object at 4500 K emits the blackbody spectrum shown here.
This object emits the most light in the __?__ part of the visible spectrum.
Answers
Answered by
DrBob222
I would think that graph which you couldn't post would give you the wavelength. At any rate, the peak is at the lower wavelengths
http://en.wikipedia.org/wiki/Black-body_radiation
I found this at 5000 K which is close to 4500 K. This shows 5,000 K about 500 or 600 nm so my guess is that 4500 K would be shifted to the red a little more and that makes me think in terms of "the red part of the spectrum" or " the upper end of the visible spectrum" or something to that effect.
https://www.google.com/search?q=black+body+radiation&client=firefox-a&hs=nog&rls=org.mozilla:en-US:official&channel=sb&tbm=isch&imgil=ajUAbSUBvHfyLM%253A%253BlYPmvkHSj9i8NM%253Bhttp%25253A%25252F%25252Fhyperphysics.phy-astr.gsu.edu%25252Fhbase%25252Fmod6.html&source=iu&pf=m&fir=ajUAbSUBvHfyLM%253A%252ClYPmvkHSj9i8NM%252C_&usg=__utlX7rQTrQQds5v5JwcUPU0vu5k%3D&biw=1024&bih=609&ved=0CEMQyjc&ei=pQd8VJ_YMMGngwT7kITYBw#facrc=_&imgdii=_&imgrc=1YHHr26oMe5MaM%253A%3BlYPmvkHSj9i8NM%3Bhttp%253A%252F%252Fhyperphysics.phy-astr.gsu.edu%252Fhbase%252Fimgmod%252Fuvcatas.gif%3Bhttp%253A%252F%252Fhyperphysics.phy-astr.gsu.edu%252Fhbase%252Fmod6.html%3B449%3B289
If you can read the wavelength from that graph, yellow is about 590 A, orange is about 650 A and red is about 700. I'll look for a chart that shows that wavelengths if you need it.
http://en.wikipedia.org/wiki/Black-body_radiation
I found this at 5000 K which is close to 4500 K. This shows 5,000 K about 500 or 600 nm so my guess is that 4500 K would be shifted to the red a little more and that makes me think in terms of "the red part of the spectrum" or " the upper end of the visible spectrum" or something to that effect.
https://www.google.com/search?q=black+body+radiation&client=firefox-a&hs=nog&rls=org.mozilla:en-US:official&channel=sb&tbm=isch&imgil=ajUAbSUBvHfyLM%253A%253BlYPmvkHSj9i8NM%253Bhttp%25253A%25252F%25252Fhyperphysics.phy-astr.gsu.edu%25252Fhbase%25252Fmod6.html&source=iu&pf=m&fir=ajUAbSUBvHfyLM%253A%252ClYPmvkHSj9i8NM%252C_&usg=__utlX7rQTrQQds5v5JwcUPU0vu5k%3D&biw=1024&bih=609&ved=0CEMQyjc&ei=pQd8VJ_YMMGngwT7kITYBw#facrc=_&imgdii=_&imgrc=1YHHr26oMe5MaM%253A%3BlYPmvkHSj9i8NM%3Bhttp%253A%252F%252Fhyperphysics.phy-astr.gsu.edu%252Fhbase%252Fimgmod%252Fuvcatas.gif%3Bhttp%253A%252F%252Fhyperphysics.phy-astr.gsu.edu%252Fhbase%252Fmod6.html%3B449%3B289
If you can read the wavelength from that graph, yellow is about 590 A, orange is about 650 A and red is about 700. I'll look for a chart that shows that wavelengths if you need it.
Answered by
DrBob222
https://www.google.com/search?q=color+vs+electromagnetic+spectrum&client=firefox-a&hs=Md1&rls=org.mozilla:en-US:official&channel=sb&tbm=isch&imgil=XYSzW09E0FFwyM%253A%253BKWT-mXLCRszi8M%253Bhttp%25253A%25252F%25252Fwww.school-for-champions.com%25252Fscience%25252Fem_spectrum.htm&source=iu&pf=m&fir=XYSzW09E0FFwyM%253A%252CKWT-mXLCRszi8M%252C_&usg=__S6wG0v4bsDVk1GEsgumAADprg9I%3D&biw=1024&bih=609&ved=0CD0Qyjc&ei=4wl8VKTTHMSYNr-kgPgM#facrc=_&imgdii=_&imgrc=XYSzW09E0FFwyM%253A%3BKWT-mXLCRszi8M%3Bhttp%253A%252F%252Fwww.school-for-champions.com%252Fscience%252Fimages%252Fem_spectrum-visible.jpg%3Bhttp%253A%252F%252Fwww.school-for-champions.com%252Fscience%252Fem_spectrum.htm%3B236%3B400
Answered by
anna
@DrBob222 Thanks! It really did help me understand it better
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