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Tamara Bojanic Chemistry Lab CH227 Lab report #6

Atomic Emission Spectra Activity Spectrum of a Single Electron Element – Hydrogen Record the line color and its position for the 3 or 4 brightest lines observed using the hydrogen lamp in table 1 and calculate each wavelength using Equation 1 from your lab manual. Be sure to include a sample of your work for one of the calculations.

Table 1: Bright Line Spectra for Elemental Hydrogen. Wavelength (nm)

Line Color

Spectral Line Position (nm)

violet

440

436

blue

490

486

red

670

666

Corrected using calibration factor

Spectrum of Multi-Electron Elements and Other Miscellaneous Spectra Qualitatively observe other atomic spectra and make notes below about the observed differences from the hydrogen spectrum. Mercury (nm): violet 445 – 4 = 441, green 560 – 4 = 556, yellow 590 – 4 = 586 Mercury and hydrogen only share violet in their spectrum. The qualitative difference in nm between the violet in hydrogen and the violet in mercury is 5 nm.

Neon (nm): yellow 590 – 4 = 586, orange 610 – 4 = 606, red 640 – 4 = 636 Neon and hydrogen share red in their spectrum. The qualitative difference in nm between the red in hydrogen and the red in neon is 30 nm.

Qualitatively compare the 3 brightest lines from the emission spectra of the elemental mercury lamp to the spectra of a fluorescent lamp and an incandescent bulb and the red and green colored sources provided. - Mercury and fluorescent spectra are not comparable, since fluorescent spectra is continuous and mercury is line spectra - Mercury and red spectra don't share any color, since the only color in red spectra is red. - The mercury and green spectra share green. There is no difference between green in mercury spectra and green in green spectra.

If daylight is available and after giving time for your eyes to adjust, observe the solar spectra by looking at the ambient daylight through a window (no worries if it is cloudy, just aim at the sky). Never aim directly at the sun. Comment, whether or not the solar spectra a continuous spectra. If not, what do you think the discontinuities represent? The spectrum appeared to be countinuous, ranging from 400 – 700 nm.

Data Analysis 1)For the first four electronic transition in the Balmer Series, calculate the change in energy of the electron (ΔE), the predicted energy of the emitted photon ( Ephoton) and the predicted wavelength of the emitted photon (λphoton). Put the calculated values in Table 2 and be sure to clearly show an example of each calculation in the space provided. Table 2: Calculated Values for the Balmer Series of Hydrogen. Electronic Transition

ΔE (J)

Ephoton (J)

Λphoton (nm)

n3 → n2

0.3025 * 10^-18

-0.3025 * 10^-18

657

n4 → n2

0.4084 * 10^-18

-0.4084 * 10^-18

486

n5 → n2

0.4574 * 10^-18

-0.4574 * 10^-18

434

n6 → n2

0.4840 * 10^-18

-0.4840 * 10^-18

410

Clearly show the following calculations for n3 → n2 transition. -- the change in energy of the electron, ΔE ΔE = 2.178 * 10^-18 J * (1/nfinal^2 – 1/ninitial^2) ΔE = 2.178 * 10^-18 J * (1/4 – 1/9) = = 0.3025 * 10^-18 J -- the predicted wavelength of the emitted photon, Λphoton ΔE = h * c / Λ Λ = h * c / ΔE Λ = 2.9979 * 10^8 m/s * 6.6261 * 10^-34 J s / 0.3025 * 10^-18 J = = 65.667 * 10^-8 = 657 nm 2) Based on your theoretical calculations, match the electronic transitions in the Balmer Series to the spectral lines you observed, and document your choices in Table 3. Then calculate the percent error between your experimentally determined and calculated wavelengths.

Table 3: Comparison of Experimental and Accepted Wavelengths from the Balmer Series. Spectral Line Experimental Λ (nm) Color Observed from Table 1

Accepted Λ (nm) from Table 2

Electronic Transition

Percent Error

red

666

657

n3 → n2

1.37%

blue

486

486

n4 → n2

0.00%

violet 436 434 n5 → n2 Below, clearly show your percent error calculation for the n3 → n2 transition.

0.46%

Percent error = (experimental – accepted) / accepted * 100% = = (666 nm – 657 nm) / 657 nm * 100% = = 9 nm / 657 nm * 100% = = 1.37 %

3) It is not impossible to observe the n7 → n2 transition in Balmer Series. Why do you think is that? ΔE = 2.178 * 10^-18 J * (1/nfinal^2 – 1/ninitial^2) ΔE = 2.178 * 10^-18 J * (1/4 – 1/49) = = 0.5 * 10^-18 J Λ = h * c / ΔE = 2.9979 * 10^8 m/s * 6.6261 * 10^-34 J s / 0.5 *10^-18 J = 397 nm This wavelength (397 nm) is not visible for humans.

4) Emission spectra are sometimes reffered to as atomic fingerprints. Is it possible to use them to identify elements in an unknown spectra? Explain your reasoning: thing about Hg spectrum and that of a fluorescent bulb. Further study of emission spectra revealed that each element produces a different set of spectral lines. Thus, the energy difference between energy levels is unique for atoms of each element. For this reason, the line spectrum of a given element is unique and is sometimes reffered to as an "atomic fingerprint".

5) Why do you think sodium vapor lights cast a different color (yellowish) than fluorescent lamps? Every element has its own unique color spectrum. You will find mercury vapor in fluorescent lights, and its ultraviolet spectrum interacts with various phosphors on the f luorescent tubeto produce artificial "white" light. Sodium's spectrum, on the other hand, is strongly yellow or orange. So, these two types or artificial lighting are vastly different in their illumination.

6) Calculate the ionization energy of the hydrogen atom. Think about this process as taking an electron from its ground state, n = 1, to a position/energy level far, far away from nucleus, n = ∞. Λ = 1 / ((1/1^2 – 1/ ∞^2) * 1.097 * 10^7 1/m) = = 1 / 1.097 * 10^7 1/m = = 9.11 * 10^-8 m ΔE = (h * c * 6.022 * 10^23) / Λ = = 1312 kJ/mol...