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kmno4
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"The CrO5 test is not easily performed here. CrO5 (actually, CrO(O2)2, the diperoxo complex of hexavalent chromium) is not easily made in your
solutions. H2O2 does not oxidize chromium(III), not even in the presence of a strong oxidizer like HNO3"
That is why I suggest to use solution no. 7
It is alkaline (strongly I suppose) and in these conditions
Cr is in Cr(OH)4 (-) form. This hydroxo comlex is easily attacked by H2O2 to give CrO4(2-).
Next step is to make it acidic (HNO3) and again add H2O2.
If Cr is present, under these conditions it will give CrO5.
It is unstable and this test should be made on cold.
Additionally, you should add (and shake with) some extrahent : diethyl ether, isoamyl or buthyl alcohol etc.. to extract CrO5.
Organic layer will be deeply blue colored if Cr is present.
Optionally you can use persulfates instead of H2O2.
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woelen
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The latter is not true. Peroxodisulfate does NOT give peroxo complexes. It reacts with hydrogen peroxide, giving oxygen, where hydrogen peroxide acts
as reductor and the peroxodisulfate acts as oxidizer.
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not_important
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Dump some Al foil bits into some of the solution, to reduce the transition metals down to lower oxidation states. Add aqueous ammonia to make
alkaline, then evaporate everything and finally heat it hot enough to drive off ammonium salts. Then check the residue for transition metals, sans
interference from excess ammonium and chloride ions.
Note the Al will give hydroxide/oxide that serves as a carrier for trace metal ppts. As example if the heated solids, after cooling, were to be
treated with a small amount of H2SO4 to dissolve the solids, then strong aq NH3 added, Ni and Cu will go into solution while Fe and other metals that
don't form NH3 complexes will remain with the Al hydroxide/oxide.
[Edited on 26-11-2010 by not_important]
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blogfast25
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Well, I carried out two tests on Cr (III) solutions, one an old Cr alum solution, one a freshly prepared Cr3+ solution (from dichromate + alcohol),
both free of iron. I got some strange results. I alkalised both strongly to dissolve as chromite. Then I added 3 % H2O2 and heated. From the first
solution Cr(OH)3 precipitated but the supernatant solution was yellowish.
The second one got the same treatment and turned… brown! To both was then added some conc. sulphuric acid and the first solution then also turned
brown:
I transferred a bit of the left hand side solution to a test tube and added a solution of Mohr’s Salt (ferrous ammonium sulphate), the solution
turns green (photo doesn't do justice to colour, the green is lighter to the naked eye):
This strongly suggests some oxidation took place during the alkaline H2O2 treatment: Cr (VI) formed oxidises Fe (II) to Fe (III) and is reduced back
to Cr (III)…
The problem with not_important’s idea is that ammonia will precipitate also the Sn, it would be nice to just gather all the D-blocks without the
Sn… But that might not matter much…
[Edited on 26-11-2010 by blogfast25]
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blogfast25
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Well, it looks like I’ve proven my own hypothesis regarding the cause of the lime green colour, namely an unknown complex involving Fe (III), Sn
(IV) and Sb, fairly conclusively wrong. Here’s what I did:
1. 0.9 g of Sn (99.9 %) and 0.1 g of Sb (99.2 %) together were given the usual HCl/HNO3 mix + NH4Cl treatment to simulate making the
hexachlorostannate from an Sb-rich pewter. Iron was supplied from the 22 % HCl. No green was ever observed, instead the solution was a strong
yellow (see below, tube 10).
2. an attempt to make NH4SbCl6 resulted in an unknown but water soluble Sb compound. Here, 5 g of Sb (99.2 %, a coarse powder) was given the same
treatment as I would pure tin but the result was different: no well formed crystals appeared and at the end of the simmering a yellow
syrupy resulted. Overnight that solidified into a hard, opaque mass, clearly looking like a hydrate, very different from (NH4)2SnCl6. Is this a (III)
or a (V) compound of Sb? Must look into that…
3. Some of the hard mass of 2. was combined with an more or less equal amount of iron free, earlier prepared (NH4)2SnCl6 and dissolved (with heat) in
about 20 ml of iron bearing HCl 22 %. It dissolves into a yellow perfectly clear liquid. Which was then reduced to just the few ml
seen below (tube 11).
The photo below summarises the results:
10: lime green control
10 and 11: solutions obtained from experiments 1. and 3.
I might corroborate this by reducing the solutions with Al and reoxidising with HNO3…
Assuming my results here are correct then it’s reasonably to assume that something else in the tankard pewter is causing the lime green colour.
Further future experiments should include working with completly Fe free HCl and adding a small amount of a Cr (III) compound to the experiments
above...
[Edited on 28-11-2010 by blogfast25]
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not_important
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Cold be both Fe and Cu, the 1st being a common contaminate and occasionally used in base metal alloys, the 2nd definitely being used in some such
alloys. The high concentrations of other ions might be messing up the tests being used.
Re my suggestion - the heating to drive off ammonium salts would likely reduce the solubility of the hydrated tin oxides, a leach with cold dilute
H2SO4 may just selectively pull ot the transition metals while leaving most Sn (and Sb) behind.
Like how you are trying to figure this one out, theory and experiment. While modern equipment might quickly give the answer, the old methods can do
the job while helping improve lab techniques too; useful for those with limited funds.
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blogfast25
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Cu had already been eliminated basically on another thread: the pewter doesn’t appear to contain any significant amounts (traces cannot be excluded,
of course).
Regarding modern equipment, trust me that if I had some ‘CSI Miami style’ stuff I’d have peeked by now! It’s only because the colour is really
unusual in inorganic chemistry that I’m still pursuing it…
The objection I have to reduction with Al followed by precipitation with strong ammonia is that you don’t really separate anything from anything:
ammonia isn’t strong enough to dissolve, at least AFAIK, Sn, Al and Cr (assuming present) as anions. So basically you’ll get a mass of everything
plus alumina that should be OK for XRF. Unless selective leaching as you suggest works. Instead I’ll reduce with Al and precipitate with strong
NaOH: at least Sn, Al and any Cr will be separated from the Fe.
The antimony compound I prepared is possibly slightly hydrolysed HSbCl6, a strong acid that, acc. Holleman, can be crystallised as HSbCl6.4.5 H2O.
This is a very strong acid, probably highly soluble in water (as consistent with observation) as indicated by the existence of a hydrate. I’ve been
preparing a second batch now, using a slightly modified procedure…
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Arthur Dent
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blogfast25 said: "trust me that if I had some ‘CSI Miami style’ stuff I’d have peeked by now!"
http://orbitingfrog.com/blog/2008/07/02/make-your-own-spectr...
and with that crude apparatus, maybe a flame test could give you info on the trace elements of your solution. 
I have some spare platinum wire if you want to make a little flame test wand... 
Robert
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blogfast25
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@ Arthur:
I’ve got one of those (but mine is better – reflector diffractors give better results than ‘see through’ diffractors).
The problem is that trace elements, even when flamed, often only emit very weak lines that are hard to see. For that you often need very hot flames, a
fairly high feed of solution into the flame or even better (but harder to do), electrical excitation (electrical arcing) of the trace element atoms.
Also you need to be able to precisely measure the wavelength of any observed lines to be able to tag them to a specific element. I’ve got blueprints
for a home made spectrogoniometer (allows determination of wavelength) in my drawer for months now but haven’t gotten round building the damn thing
yet. And the instrument doesn’t solve the problem of sufficient atom excitation either.
So, dear Watson, it’s not an open and shut case…
BTW, if you want to have a go at the spectroscope you linked to, don’t use ‘an old CD’ (that’s yesterday’s homemade spectroscope technology
– LOL!), instead use an R-DVD: the datalines on R-DVDs are much closer together leading to much better resolution (about 1 nm for a good build).
Mine uses an R-DVD reflector diffractor…
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Arthur Dent
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Thanks for the tip about DVDs, I didn't know that. I do know that the very small plastic-encased lens right in front of the laser diode in some CD/DVD
drives (like the PHR803) is in fact a diffraction grating. Unfortunately, it's about 1.5mm x 1.5mm, too small for this particular application. This
site:
http://www.rainbowsymphonystore.com/difgratfilsh.html
sells diffraction gratings in flexible sheets, and the price is right!
But definitely, I'll try to use an old DVD for this project.
One of my "other" hobbies is playing around with lasers and laser optics, still hoping to get my hands on a nice (cheap) multiline argon laser some
day. 
Robert
[Edited on 29-11-2010 by Arthur Dent]
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blogfast25
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Well, well, well: chromium is back on the top of the list of suspects with a simple experiment, similar to the one above. Tin filings (a ‘pinch’)
and Cr filings (from aluminothermic chromium – a smaller ‘pinch’) were combined with 15 ml of HCl 22 % and 5 ml of HNO3 38 %. Reaction starts
immediately and a lime green colour appears.
The immediate appearance is a slight difference with the green appearing when dissolving pewter in ‘aqua regia’: there it appears much later on.
But the physical form of the Cr in this test is also very different.
The solution (appears a little darker than to the naked eye):
A small amount of the solution was then mixed with yellow HCl 22 % and it was possible to obtain a perfect colour match: see 1 (control) and 12
(matched solution):
In itself this still doesn’t prove anything conclusively and only proving the pewter actually contains Cr will do. But it does quite fit the facts:
due to the concentrating of pewter AR solutions to crystallise out the hexachlorostannate, as well as the concentrating of an iron bearing HCl
solution during the same step, relatively little Cr would be needed for the yellow FeCl3 and green CrCl3 to combine to a lime green colour.
O/T:
It’s that cold here that despite some electrical heating a saturated solution of oxalic acid on my shelves started to crystallise out! California it
ain’t…
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blogfast25
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Quote: Originally posted by Arthur Dent  | Thanks for the tip about DVDs, I didn't know that. I do know that the very small plastic-encased lens right in front of the laser diode in some CD/DVD
drives (like the PHR803) is in fact a diffraction grating. Unfortunately, it's about 1.5mm x 1.5mm, too small for this particular application. This
site:
http://www.rainbowsymphonystore.com/difgratfilsh.html
sells diffraction gratings in flexible sheets, and the price is right!
But definitely, I'll try to use an old DVD for this project.
One of my "other" hobbies is playing around with lasers and laser optics, still hoping to get my hands on a nice (cheap) multiline argon laser some
day.
Robert
[Edited on 29-11-2010 by Arthur Dent] |
Robert:
The place to be for homemade spectroscopy is here (click around, it’s full of useful info for a beginning spectroscopist!):
http://astro.u-strasbg.fr/~koppen/spectro/mk2be.html
That’s the model (well, three of them!) I built using an R-DVD. It resolves the famous sodium D doublet (on a good day!) If set up and used properly
emission lines are very clear and narrow. But for serious spectroscopy you need good excitation and a goniometer to measure angles and thus calculate
wavelengths. My planned instrument will measure angels down to 0.1 (360) degree. That’s about 1 nm. Commercial instruments like that should not cost
much more than about 200 bucks I’m guessing.
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Arthur Dent
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Shweeet!
I guess I have my work cut out this weekend, really cool project! I happen to have a nice black cardboard box just right for the job.
Glad the mystery trace impurity got identified!
- Robert
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not_important
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If you can co-display, as two bands, the sample spectrum and that of a neon light - cost around a half Euro, you can do a decent job of wavelength
identification. Neon hass a number of lines throughout much of the visible and so makes a decent low cost reference.
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blogfast25
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@not_important:
You mean a baby room night light or something like that? I've used the Na D line, Li scarlet line and a 617 nm red laser as references. Never thought
of Ne though...
Back to the job at hand: Cr in the lime green solution. Despite a rather elaborate effort I've still not been able to conclusively prove the green is
that of Cr3+.
50 ml of the solution was neutralised with NaOH, to as close as possible as pH = 7. This solution is a light sky blue. Filtered and dried the
precipitate (mainly SnO2) till it was a paste. About 1 g of it was mixed with an excess of Na2CO3.10H2O and quite a bit of NaNO3. The crystal water
was first removed by gently heating the mix in a ceramic crucible and the dry was mix then fired at max heat for 30 mins (propane Bunsen). The bottom
of the crucible was glowing bright red-orange.
After cooling the frit showed a yellow tinge at the bottom and was removed with small aliquots of 50 % H2SO4 (quite a bit of NO2 was surprisingly
generated at that stage…) The slurry (the SnO2 remains as solid) was then filtered.
But the filtrate didn’t test positive for Cr: with H2O2 (3 %) the colour changes to yellow, adding lead nitrate to it yields a white precipitate,
not yellow lead chromate. Adding lead nitrate to the solution containing peroxide also gave a white precipitate.
Tomorrow a test with NaOH – KClO3… and a higher quantity of the precipitate.
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not_important
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Neon bulb, plus a series resistor. Bulb like this : http://www.weisd.com/store2/EIKA1A.php
sample spectra including Ne here http://www.amateurspectroscopy.com/color-spectra-of-chemical... notce that there are strong lines for most regions of the visible.
Someone else mentions the concept here http://chemcom.tripod.com/cTakingSpectraPictures.htm
I'd be surprised if there's Cr in that alloy, as that doesn't seem to be common, but if the tests indicate it is there then I'll accept that. Perhaps
the Cr got in through salvaged metal, a small amount of Cr such as from plating on metal or plastic could dissolve in the alloy and as it's not lead
get past normal tests.
Does the white precipitate with lead nitrate match the characteristics of the chloride or sulfate? (check solubility in hot water) Also boil the ppt
with Na2CO3 solution, filter off lead, check solution for other complex anions.
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kmno4
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| Quote: Originally posted by blogfast25 |
After cooling the frit showed a yellow tinge at the bottom and was removed with small aliquots of 50 % H2SO4 (quite a bit of NO2 was surprisingly
generated at that stage…)
|
In this way you probably reduced all, or almost all hypotetical chromates.
| Quote: |
But the filtrate didn’t test positive for Cr: with H2O2 (3 %) the colour changes to yellow |
So what was the colour of the filtrate ?
blogfast25 - I think you should exercise converting Cr(III) to Cr(VI) via NaOH/H2O2 methode and convertig it further to CrO5.
I did this several times and it always works if done properly.
If you have already Cr in solution then all "hot methodes" are simply waste of time.
If content of Cr is small - you possibly will not see any blue colour of CrO5 in such diluted solution. Then you must use extrahent for CrO5: to
acidic chromate (max 0,1 M) at 10 C (or less) add ~1cm3 of diethyl ether or butanol (as far I remember butyl acetate will not give good results) and
next add H2O2 (excess is better than too small amount). Shake it at once - if organic layer is blue, then Cr is present. If not - most probably it is
not present.
If there is no colour then you can add small amount of any di/chromate (soluble) and see if blue colour will really appear. If yes, then conditions
are proper. If not.... something is wrong.
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blogfast25
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@ kmno4:
OK, back to the drawing board.
The control solution in a test tube was first alkalised with 5 M NaOH: SnO2 precipitates from it, then redissolves to weak green solution (stannate
and chromite). Weak H2O2 added: colour change to yellow-orange is very marked, this clearly indicates chromate formation. Then re-acidified with HNO3
38 %: no notable change (pH is now < 1). Added Pb(NO3)2, alas no yellow lead chromate forms, instead some PbCl2 slowly dropped out.
To half of the acidified solution more H2O2 was then added, no change occurred though…
I believe to see PbCrO4, I’d have to convert all to nitrates, washing out any chlorides, then re-apply the procedure…
There’s another elephant in the room that everyone seems to have missed, even kmno4 (;-)): MnO4(-) according the reduction potential series should
be capable of comfortably oxidising Cr (III) to (VI). Titration of a test tube of the lime green solution using 0.1 N KMnO4 confirms that: the MnO4(-)
reacts away swiftly and the green colour of Cr3+ disappears… Could even be the basis of a volumetric determination of the Cr...
So it must be chromium alright. I’m 100 % sure, as the early tests confirm, that it’s part of the alloy and didn’t come flying in from a cutting
tool or post-digestion. Weird…
I’ve another piece of pewter, an old style hip flask (ideal for the drink-driving ‘hip flask defence’! – ‘I just took a big swig from my
steeping alcohol filled hip flask when getting out of the car, officer, you’ve got nothing on me!’), I’ll see if that contains any soon…
[Edited on 1-12-2010 by blogfast25]
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blogfast25
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@ not_important:
Like the neon bulb but its wattage is seriously low. Thanks for the interesting links…
The one about digital photography of spectra is very useful: these things are really seriously difficult to capture well. Here’s my best shot of a
saver bulb spectrum with an R-DVD based spectroscope with an 8 MPixel camera:
To the naked eye the lines are much finer.
Where the link goes a bit off the rails is where he writes about the ‘calibration scale’. I tried several times something similar with very
conflicting results until thick as two short planks here remembered an experiment carried out a few weeks before for the determination of the
wavelength of a pointer-style red laser. Below is shown the principle of the method: Top: a diffraction grating (a CD and subsequently also an R-DVD).
Left: path of incident ray with laser in left hand bottom corner, right path of reflected/diffracted ray for order n=0 and n=1 (this is a real life
situation):
The wavelength was determined from (a set of measurements with varying beta) the angles alfa and beta, in turn calculated with Pythagoras from the
lengths O’O, O’B and O’C. It doesn’t take much to understand that the scale obtained is highly non-linear. With photos the situation is even
more complicated because the camera plane and the reflector/diffractor plane are unlikely to be exactly parallel…
The only real way to measure the angles easily is by means of a Bunsen-Kirchoff style ‘swivel’ spectrometer but with a diffractor instead of a
prism (prisms still work too of course…).
So the author’s points on calibration seems a mix of wishful thinking and confirmation bias to me…
[Edited on 1-12-2010 by blogfast25]
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watson.fawkes
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| Quote: Originally posted by blogfast25 | With photos the situation is even more complicated because the camera plane and the reflector/diffractor plane are unlikely to be exactly
parallel…[...]
So the author’s points on calibration seems a mix of wishful thinking and confirmation bias to me… |
I
don't see any fundamental reason why it can't be made to work, with appropriate caution. I should note that I didn't see on that page anywhere he was trying to do calculations of wavelength, nor did he make any claiming of linearity with the picture of the ruler. I think you
may be reading something into the narrative, which seemed to me just about image capture.
What I think is quite clever about the idea is that you can photograph quite weak intensities with this scheme by means of long exposure times.
Getting a picture of the scale itself can be done with a short flash of a white LED at some point during the exposure. By manipulating these two
times, you can get both adequate signal and adequate contrast.
As you point out, the scale will be nonlinear and require calibration. Shooting an NE-2 lamp spectrum, though, can provide that each time the device
is used. You need to know the angle from the principal axis to the scale (canonically 90°<!-- --> and the distance from the optic to the scale. Three lines of known wavelength suffice. If you don't know the spectrum
grating spacing, you'll need a fourth line. You might want to measure it anyway, since it has a non-zero temperature coefficient. The law of cosines
gives slightly easier arithmetic, FWIW, although you've likely more-or-less already derived it already.
As for the camera's orientation to the scale, that won't matter if you're correlating the spectral lines to the rulings on the scale directly.
Coincidence there is what you're looking for. If you feel the need to interpolate, you might have to determine distance and orientation from camera to
scale, but the easiest way to deal with this is to mechanically lock the scale at right angles to the axis of the optic, look only at the foot of the
perpendicular, ensure adequate distance, and then rely on the sinθ ~ θ small-angle approximation. In other words, by constructing
the apparatus properly, you can assume that your image is linear to the scale. And this is just for interpolation, recall, so any error introduced
here is second-order.
As to the quality of your images, I suspect vibration. Human vision compensates for vibration quite well. Mechanical systems need care. It's one of
the reasons for fancy and expensive optical tables.
Lastly: Traditional spectroscopy was always done in very dark rooms.
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blogfast25
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@ watson:
Yes, I’ve always set mine to the longest exposure times my camera allowed.
As regards the calculation you suggest, it’s principally exactly what I used for the measurement of the wavelength of the laser.
Re. the non-zero temperature coefficient, I didn’t think of that: R-DVDs and CDs are made of some plastic or other, so quite susceptible to dilation
by temperature.
I used a tripod and set timer (the camera takes a specified number of shots without operator intervention) to avoid all camera shake: I think the main
source of line widening may simple be due to limited resolution of the camera.
Dark rooms? Yes of course: not so easy to achieve when you have a de facto light source in there…
A few facts:
The actual wavelength, using the definitions used in the photo above is calculated from:
n λ = D ( sin β - sin α )
with n the order of the spectrum and D the distance between adjacent lines on the diffraction grating.
For ordinary CD ROMs, D = 1.6 μm (micrometer) and for DVD-R (single layer, 4.7 GByte storage capacity) it is 0.74 μm.
The laser light measured was 659 nm (with a sample standard deviation of 9 nm).
[Edited on 1-12-2010 by blogfast25]
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watson.fawkes
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| Quote: Originally posted by blogfast25 | | I used a tripod and set timer (the camera takes a specified number of shots without operator intervention) to avoid all camera shake: I think the main
source of line widening may simple be due to limited resolution of the camera. |
There's a remarkable amount
of vibration in an ordinary room unless you take special precautions.
There's also, I see now, the possibility that there's a diffractive spreading going on having to do with one of the restricted pupils in the camera.
What kind of camera is it?
The imaging chip is also a candidate for strange monkey business, like internal side scattering that's ordinarily never a factor in snapshots, where
there aren't huge order-of-magnitude jumps in intensity from pixel to pixel.
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blogfast25
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Going back to the lead chromate, I now believe it didn’t form because the conditions weren’t favourable at such low pH. The solubility product of
PbCrO4 is Ks = 2.8 x E-13, so not massively small (a bit unusual for something that occurs as a mineral) and low pH pushes the CrO4(2-) to HCrO4(-)
and Cr2O7(2-): there probably wasn’t enough CrO4(2-) at pH < 1 to exceed Ks…
Re the camera, it’s a box standard Canon PowerShot A560, zoom 4X, 7.1 MPixels. You can choose from a number of preset settings, such as exposure
time (on a relative scale). Decent but not professional and certainly not designed for shooting spectra! Basically an ‘instant’ digital camera
with some extra options. I use it for all pix on this forum…
What’s even stranger is that I started off using a proper non-digital camera of very high quality: a semi-professional Zeiss Ikon, about 60 years
old but in perfect condition). It’s one of those where the viewfinder looks right through the lens, until you snap and the mirror quickly veers away
briefly, exposing the analog film. Just about everything can be set as required by the operator.
Well, I set up my camera, could see the spectra perfectly (as to the naked eye), took some twenty shots varying the settings and when the film was
developed… nothing, nada, zilch! Not a single line… I used box standard Kodak middle of the range ASA film for that. Explain that to me!
I was also wondering what’s the lower limit of D for a VIS diffraction grating, the R-DVD is 740 nm. There’s a point where D is so small that
light waves basically ‘wash’ over the lines, like sea waves over a small object or light waves over an atom or simple molecule… But before that
point is reached, smaller D leads to better resolution, see CD v. R-DVD…
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watson.fawkes
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| Quote: Originally posted by blogfast25 | | What’s even stranger is that I started off using a proper non-digital camera of very high quality: a semi-professional Zeiss Ikon, about 60 years
old but in perfect condition). [...] Well, I set up my camera, could see the spectra perfectly (as to the naked eye), took some twenty shots varying
the settings and when the film was developed… nothing, nada, zilch! Not a single line… I used box standard Kodak middle of the range ASA film for
that. Explain that to me! |
The dynamic range of the eye is about 90 dB, so it's certainly possible
that the film remained underexposed. What was the longest exposure time? My immediate intuition is that you'd need an external exposure timer, but
that's only a guess.
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blogfast25
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Longest exposure time a few seconds. Yes, to an exposure timer. Film with higher light sensitivity should also help. Still, digital has such
tremendous advantages…
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