6 - 4. (1 point) If E = 186 kJ/mol, then = 643 nm . 5. (2...

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Dr. S. Berniolles, Page 1 of 1 CHEM 6BL PRELAB/POSTLAB ANSWER KEY EXPERIMENT #6: ATOMIC SPECTRA PRELAB QUESTIONS (10 points). NO POSTLAB . PRELAB ANSWER KEY 1. (2 points) Complete the table below (take to 3 sig. figs): Color: Red Green Wavelength ( λ ) 700. nm 520. nm Energy (E) 2.84 × 10 -19 J 3.82 × 10 -19 J Frequency ( ν ) 4.28 × 10 14 Hz 5.76 × 10 14 Hz The color of light with the higher: Energy (E) is: RED GREEN Frequency ( ν ) is: RED GREEN 2. (0.5 point) The color of light that boron emits at 518 nm is GREEN . 3. (2 points) If E = 350. kJ/mol, then λ = 342 nm . Could UV lamps be used to break these bonds? YES NO WHY? 342 nm is in the UV (ultraviolet), so any UV lamp emitting at <342 nm wavelength could break these C-C single bonds.
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Unformatted text preview: 4. (1 point) If E = 186 kJ/mol, then = 643 nm . 5. (2 points) The frequency, wavelength and energy (per mole) of light emitted when an electron drops from n = 4 to n = 2 energy state in Hydrogen is calculated to be: = 6.17 10 14 Hz = 486 nm (486 10-9 m) Is this emission in the visible range of the electromagnetic spectrum? YES NO E = 246 kJ/mol 6. (1 point) Convert the ionization energy of Helium (per atom) to ionization energy (per mole): E = 2,370 kJ/mol . 7. (1.5 points) If = 85.0 nm (85.0 10-9 m), then E = 1,410 kJ/mol . Is this transition larger than the I.E. of hydrogen? YES NO...
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This note was uploaded on 05/31/2009 for the course CHEM 6BL taught by Professor Berniolles during the Spring '08 term at UCSD.

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