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Unformatted text preview: Version PREVIEW Exam 1 JOHNSON (53755) 1 This printout should have 30 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. LDE Planck relation 001 001 10.0 points What is the energy, in Joules, of a photon of wavelength 200 nm? What bond energy would this correspond to, in kJ mole 1 ? 1. 1 . 32 10 40 J; 7.95 10 20 kJ mole 1 2. 9 . 94 10 17 J; 1.65 10 43 kJ mole 1 3. 9 . 94 10 19 J; 599 kJ mole 1 correct 4. 1 . 32 10 21 J; 795 kJ mole 1 5. 9 . 94 10 21 J; 5.99 kJ mole 1 6. 1 . 32 10 31 J; 7.95 10 11 kJ mole 1 Explanation: = 200 nm c = 3 10 8 m/s h = 6 . 626 10 34 J s. For a photon c = , so E = h = h c where c is the speed of light and h is Plancks constant. E = h c = ( 6 . 63 10 34 J s ) ( 3 10 8 m / s ) 1 200 nm 10 9 nm 1 m = 9 . 94 10 19 J 9.94 10 19 J 6 . 022 10 23 mol 1 = 599 kJ mole 1 LDE Classical Failure 001 002 10.0 points Which of the following statement(s) is/are true? I) Classical mechanics accurately predicted the behavior of blackbody radiators. II) The failure of classical mechanics to pre dict the behavior of blackbody radiators is called the ultraviolet catastrophe. III) A minimum frequency of light is required to eject an electron from a metal surface. IV) The emission spectra of gases are con tinuous rather than discrete. 1. I, II and IV 2. II, III, and IV 3. II and III correct 4. III and IV 5. I and III Explanation: Classical mechanics predicted that the power radiated by a blackbody radiator would be proportional to the square of the fre quency at which it emitted radiation, and thus approach infinity as the frequency in creased. This was false, since at higher fre quencies blackbody radiators emit less, not more power. This was termed the ultraviolet catastrophe. Classical mechanics also pre dicted that the energy (velocity) of electrons emitted from a metal surface is proportional to the intensity of light. In reality, the en ergy (velocity) is only dependent upon the frequency of light. Once the threshold fre quency is reached, however, the number of emitted electrons is proportional to the in tensity of light. Classical mechanics also fails in explaining the discrete lines in ab sorption/emission spectrum, which are due to discrete energy levels of atoms. LDE Rydberg Calc 001 003 10.0 points For the Hydrogen atom, an electron mov ing between which two energy levels in the Balmer series would have to absorb light with a frequency ( ) of 6.19 10 14 Hz? 1. n = 2 and n = 3 2. n = 1 and n = 3 Version PREVIEW Exam 1 JOHNSON (53755) 2 3. n = 3 and n = 5 4. n = 2 and n = 4 correct 5. n = 1 and n = 4 6. n = 1 and n = 2 Explanation: A frequency of 6.19 10 14 Hz corresponds to a transition from n = 2 to n = 4 using the Rydberg equation. The fact that the Balmer series is mentioned also indicates that n...
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This note was uploaded on 03/04/2011 for the course CH 302 taught by Professor Holcombe during the Spring '07 term at University of Texas at Austin.
 Spring '07
 Holcombe

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