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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.
- Spring '08