posted by Kim on .
I have no idea What to do. What does it mean by 'draw what you think the visible range spectrum of this dye would look like.' To get the A, I used Beer's law. So I got 9.265nm for Absorbance. What should I do for the next?
The question is
----Next Simon checks the spectrophotometer. He takes out his traveling standard solution - a 2.50 x 10-5 M solution of a particularly stable dye It has a molar absorptivity of 3.85 x 105 at 585 nm. What value for absorbance at 585 nm does Simon expect to get? Draw what you think the visible range spectrum of this dye would look like.
How did you get 9.265 nm for absorbance? A nm is a unit of length, not absorbance?
For the next part,
A = k*c (I've omitted the cell length since that usually is a constant value).
A = molar absorptivity x concn(in M)
A = 3.85 x 10^5 x 2.50 x 10^-5 = ?? That is the absorbance Simon expects.That may be how you arrived at the 9.625 in the first part of your question. The drawing part is this.
Since the solution is being measured at 585 nm, one expects that will be the point at which A is a maximum. So you have a sheet of graph paper, label the Y axis as A units and the x axis as wavelength and show 400 nm in the left side and 700 nm on the right side. Then place a point at 585. That is where the maximum will be. Now draw a smooth curve, beginning at 400 nm and zero A to the point and back down to zero A at 700 nm (UNLESS you have some other information that gives different A values for other wavelengths. Here is a sample spectrum of how a visible spectrum should appear but the maximum is not at 585.
Then, What should be the unit for 9.625? I thought that is the absorbance. So I used nm unit which is for absorbance.
You need to work out the unit by including units in the calculation.
9.265 cannot be an absorbance as the scale on even the best instruments only goes to just above 3.
585 nm is a standard filter value so Simon dye sample may be such it has a maximum at 585 nm.
Absorbance itself does not have any units as it is a ratio of two log values. This is done by the instrument and is invisible to the operator. Only folks like me who built their own spectrometer had to do this by hand!