Christina Svensson Mr. Porter SCH3UE- 03 September 9th 2008 Flame test and Spectroscopy (emission line spectra)
Part A) Spectroscopy lab: 1) Identifying the unknown gasses: The emission spectra for the unknown gasses match with following gasses: Unknown #1: Mercury Unknown #2: Helium • •
Element (gas): Hydrogen
Spectrum:
430-440
490-500
690-700
Mercury
400-410
440-450
560-570
600-610
Argon
430
500
630
690
730
580
620
690
690
720
Crypton
430-460 Helium
390-410
450-460 480490
510 530
520590-610
Neon
550
560
~600
700 650 750
720 740
Unknown #A
410-420
440-450
560-570
390-400
450-460 480490
600-610
Unknown #B
500-510 520530
600-610
700
730
2) Calculate the energy associated with each of the spectral lines visible for hydrogen: To figure out the energy of each visible spectral line, we must calculate the amount of energy for each energy level, since the formula for calculating energy for the visible spectral lines is: Ecolour = Efinal level-Einitial level
Sample calculation I: En= -13,6 eV/n2
E2= -3,4 eV
E2= 13,6 eV/22
E3= -1,5 eV
E2 = -13,6 eV/4
E4= -0,85 eV
E2 = -3,4 eV Sample calculation II:
E5= -0,54 eV E6= -0,38 eV 1
Ecolour= Efinal level-Einitial level
Ered=1,9 eV
Ered= E3-E4
Eturquoise= 2,6
Ered=(-1,5eV)-(-3,4eV)
Eviolet(a)= 2,9 eV
Ered = 1,9 eV
Eviolet(b)= 3,0 eV
Since we know that the colour with the shortest wavelength is violet, and the shortes wavelength equals the highest energy, than the colour with the highest e nergy must be violet.
This calculation can also be done using E= hc/λ 3) Scientists have decided that the energy of E ∞= 0 and any lower energy would be negative, meaning the energy of E ∞ equals to 0. using the given formula: En= -13,6 eV/n2
E∞= -13,6 eV/ ∞ 2 And since the denominator is infinitely large, the energy= 0. 4) The light produced by a sodium vapour lamp shows two lines with wavelengths 5.89 x 10-7 m and 5.90 x 10 -7 m. Identify the colour of the sodium vapour lamp lamp using the data collected from the experiment. As we saw in experiment experi ment A, the flame lab, the sodium turned yellow after burning off the alcohol. And according to the info we collected for the spectroscopy lab, the colour most likely to be about 589-590, is yellow. Part B) Flame test:
Observation table: Element (salt)
Name of element
Qualitative observations
Colour of flame (after the alcohol burned of)
LiCl
Lithium Chlorine
White powder
Bright magenta/deeper red
SrCl2
Strontium chloride
White powder
Red flame wrapping in blue flame
CuCl2
Copper(II) chloride
Blue powder
NaCl2
Sodium Chlorine
White powder
Turquoise with the tip of the flame being red Orange yellow
BaCl2
Barium chloride
White powder
Pale green
Cl
Chlorine
White powder
Purple
CaCl2
Calcium chloride
White powder
Orange
This means that the Chlorine flame releases the most energy by having the shortest wavelength (purple). And the Calcium Chloride, Sodium Chloride and Strontium Chloride
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have the longest wavelengths, and therefore have the smallest amount of energy being emitted.
Wrap up:
Using the equation:
E= hc/λ Where •
• • •
E= energy c= speed of light: 3,00 x 10^8 m/s h= Planck’s Constant: 6,63 x 10^-34 Js(Joule seconds) λ= wavelengths
For λ= 400 nm: E= hc/λ E= (6,63 x 10 -34 Js)(3,00 x 10 8 m/s)/400 nm E= 1,99 x 10 -25 Js/4,00 x 10 -7 m E=4,96 x 10 -19 J For λ=700 =70 0 nm: nm : E= hc/λ E= (6,63 x 10 -34 Js)(3,00 x 10 8 m/s)/700 nm E= 1,99 x 10 -25 Js/7,00 x 10 -7 m E=2,84 x 10 -19 J And that is why the visible spectra goes from E=2,84 x 10 -19 J to E=4,96 x 10 -19 J.
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