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jbebo01 jbebo01
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Posts: 25
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10 years ago
1.Beta-carotene absorbs primarily blue and violet light in the visible spectrum. 
Explain why it appears orange.  Do you see the light that is absorbed?
What do you see instead

2.Molecular vibrational transitions absorb infrared radiation, while electronic transitions of double bonds absorb ultraviolet radiation. Use Figure 7.5 in Ebbing, along with the equations E=hc/ and E=h, to explain how the energies of infrared and ultraviolet compare with visible light transitions. 
That is, do they have higher or lower energies than visible light?


3.Suppose an atom absorbs a photon and undergoes a quantized transition from a lower energy level to a higher energy level.  Explain how you would mathematically determine the energy and frequency of the photon.  That is, explain the concept of  E for the two energy levels.
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Staff Member
Educator
10 years ago
Do you see the light that is absorbed?

You don't see the light that is absorbed. For example, if you see something that is green, it has absorbed all other wavelengths except for the wavelength that is visually green.

Sorry, jbebo01, this is all I know.
Mastering in Nutritional Biology
Tralalalala Slight Smile
wrote...
Subject Expert
10 years ago
Hi jbebo01,

1. As padre said, you don't see what's being absorbed, only what's not absorbed. In other word, you see the complement color of what's being absorbed. The complement color of Blue is Orange and of Violet is Yellow, so you would see something between yellow and orange.

2. E = hf  (f is frequency)        c= wavelength* frequency
The higher the frequency ( or the lower the wavelength), the higher energy the light has.
Therefore, violet > visible light > infrared, energy-wise.

3. Let E = energy absorb, f = frequency, h = Planck's constant, a = 1st energy level, b = 2nd energy level
Equations:
E = 2.18x10^-18* (1/b^2 - 1/a^2)
E = h*f
Rightwards Arrow  2.18x10^-18* (1/b^2 - 1/a^2) = E = h*f
 2.18x10^-18* (1/b^2 - 1/a^2) = h*f
( 2.18x10^-18* (1/b^2 - 1/a^2))/h  = f

Hope this helps,
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