# 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.

## 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.

## 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

## @DrBob222 Thanks! It really did help me understand it better

## To determine the part of the visible spectrum where the object emits the most light, we need to understand the concept of blackbody radiation and how it relates to temperature.

Blackbody radiation refers to the spectrum of light emitted by an object that absorbs all incoming radiation and re-emits it. The spectrum of blackbody radiation depends on the temperature of the object. As the temperature increases, the spectrum shifts towards shorter wavelengths (higher energy photons).

To find the part of the visible spectrum where the object emits the most light, we can use Wien's displacement law. The law states that the wavelength at which the intensity of radiation emitted by a blackbody is maximum is inversely proportional to its temperature.

Mathematically, Wien's displacement law is expressed as:

λ_max = b / T,

where λ_max is the wavelength where the intensity is maximum, b is a constant known as Wien's displacement constant (approximately equal to 2.898 × 10^-3 m⋅K), and T is the temperature of the object in Kelvin.

In the given question, the temperature of the object is 4500K. We can plug this temperature value into the equation to find the wavelength, λ_max.

λ_max = (2.898 × 10^-3 m⋅K) / (4500K)

Calculating this equation gives us:

λ_max ≈ 6.44 × 10^-7 meters.

Now, to determine the part of the visible spectrum where the object emits the most light, we need to convert this wavelength to a more familiar unit, such as nanometers (nm), which is commonly used when discussing the visible spectrum.

1 meter = 1 × 10^9 nm

Therefore, the wavelength in nanometers can be calculated as:

(6.44 × 10^-7 meters) × (1 × 10^9 nm / 1 meter)

Calculating this equation gives us:

λ_max ≈ 644 nm

So, the object emits the most light in the part of the visible spectrum around 644 nanometers. This corresponds to the red region of the spectrum.