Christina SvenssonMr. 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): Spectrum:          Hydrogen              430-440   490-500     690-700Mercury              400-410 440-450   560-570 600-610  Argon                430 500 630 690 730Crypton                430-460   580 620 690Helium            

  390-410450-460 480-490

510 520-530 590-610 690 720

Neon            

      550 560 ~600 650700 720 740 750

Unknown #A              410-420 440-450   560-570 600-610  Unknown #B            

  390-400450-460 480-490

500-510 520-530 600-610 700 730

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:

Sample calculation II:

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E2= -3,4 eV

E3= -1,5 eV

E4= -0,85 eV

E5= -0,54 eV

E6= -0,38 eV

En= -13,6 eV/n2

E2= 13,6 eV/22

E2 = -13,6 eV/4

E2 = -3,4 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 energy 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 using the data collected from the experiment.

As we saw in experiment 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

observationsColour 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 Turquoise with the tip of the flame being red

NaCl2 Sodium Chlorine White powder 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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Ecolour= Efinal level-Einitial level

Ered= E3-E4

Ered=(-1,5eV)-(-3,4eV)

Ered = 1,9 eV

Ered=1,9 eV

Eturquoise= 2,6

Eviolet(a)= 2,9 eV

Eviolet(b)= 3,0 eV

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/λE= (6,63 x 10-34 Js)(3,00 x 108 m/s)/400 nmE= 1,99 x 10-25 Js/4,00 x 10-7 mE=4,96 x 10-19 J

For λ =700 nm: E= hc/λE= (6,63 x 10-34 Js)(3,00 x 108 m/s)/700 nmE= 1,99 x 10-25 Js/7,00 x 10-7 mE=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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