Question

Please help with this chemistry lab question

ATOMIC EMISSION SPECTROSCOPY ADVANCED STUDY ASSIGNMENT Name Section Date This assignment MUST be completed before the laboratory period begins. Wavelength values for the hydrogen emission spectrum are required for the experiment. A few answers have been provided for you. You may wish to verify some of these to ensure that your calculation methods and signifcant digits are correct. 1. Calculate energy values for the first SIX hydrogen energy levels usings En =-RH/n--(2.179 x10-18J/n2 where n = 1,2,3,4 6 Express your answers in 10-2 exponential form as a convenience for later calculations. -217.9o x 10-20J -13.618 × 10-20J x 10-20J x 10-20 E, x 10-20 -6.0527 × 10-20J 2. Calculate all possible energy level DIFFERENCES between Level 1 and levels 2 through 6 Level 2 and levels 3 through 6 Level 3 and levels 4 through 6 a. b. SUMMARIZE the ΔΕ values in the table below c. 3. Calculate the corresponding wavelength for each possible ΔE value using: A- hc/AE (6.626 x 10J.s)(2.998 x 101 nm s-1/AE SUMMARIZE the wavelengths on the table below underneath their corresponding ΔΕ. igher n lower 211.8, x 10-20J 6 x 102 x 10-20J x 10 204.2, x 10-20 J 97.24, nm 121.5, nm nm nm nm x 10-20J 30.26, x 10-20J 656.3s nm 48.42, x 10-20J x 10-20 410.2, nm434.1,nm nm 18.15 x 10-20 15.49, x 10-20J 10.59, x 10-20] 1093.g nm nm nm Diagram each energy level (E, thru Es) on the reverse. Diagram all possible transitions between levels. Label each transition with its wavelength. Identify the ultra-violet, the visible, and the infrared wavelengths. 4. 63

Answer 1

1. The energy of the energy levels are

The energy of the energy levels are E_n = -R_H/n^2 = -(2.179 times 10^- 18 J/n^2 where n = 1, 2, 3, 4....6 Express your answers in 10^- 20 exponential form as a convenience for later calculations E_1 = -217.9_0 times 10^- 20 J E_2 = -54.48 times 10^- 20 J E_3 = -24.21 times 10^- 20 J E_4 = -13.61_8 times 10^- 20 J E_5 = -8.716 times 10^- 20 J E_6 = -6.052_7 times 10^- 20 J a) The energy differences can be calculated as follows. delta E = E_2 - E_1 = [-54.48 -(-217.9)] x 10 - 20 J = 163.42 x 10^-20 J delta E = E_3 - E_1 = [- 24.21 -(- 217.9)] x 10 - 20 J = 193.69 x 10^-20 J delta E = E_4 - E_1 = [-13.61 -(-217.9)] x 10-20 J = 204.29 x 10^-20 J

2. a) The energy differences can be calculated as follows.

$\Delta$E = E₂ - E₁ = [-54.48 -(-217.9) ] x 10^-20 J = 163.42 x 10^-20 J

$\Delta$E = E3 - E₁ = [-24.21 -(-217.9) ] x 10^-20 J = 193.69 x 10^-20 J

$\Delta$E = E₄ - E₁ = [-13.61 -(-217.9) ] x 10^-20 J = 204.29 x 10^-20 J

$\Delta$E = E₅ - E₁ = [-8.716 -(-217.9) ] x 10^-20 J = 209.184 x 10^-20 J

$\Delta$E = E₆ - E₁ = [-6.052 -(-217.9) ] x 10^-20 J = 211.848 x 10^-20 J

Similarly you can calculate for 2.b) and 2.c)

3. using above answers you can calculate the values in the table.

If you get any doubt please comment. I shall help you.

4. The spectrum is as follows

*****Please rate*****Thumps up*****