Physics 30 Jan 01 Diploma Exam

Under most circumstances 025 mev per nucleus is

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Unformatted text preview: temperature of the plasma, in Kelvin, and b is a physical constant equal to 1.4 × 10–23 J/K. One method of obtaining the temperatures necessary for fusion is to use a high-intensity laser to heat a small cluster of nuclei. One such laser emits a 1.0 × 1015 W pulse of ultraviolet radiation that lasts for 1.0 × 10–12 s. The wavelength of this laser is 280 nm. A Fusion Reaction Equation 2 1H 31. + 3H → 1 X + neutron The missing product, X, in the fusion reaction given above is A. 5 2 He B. 4 2 He C. 4 1H D. 3 2 He 20 20 32. The main reason that the nuclei need to have such large kinetic energies is that A. B. C. D. 33. When the average kinetic energy of the nuclei in a plasma is 0.25 MeV, then the temperature is A. B. C. D. 34. fusion releases large amounts of energy fission must occur before fusion can occur this kinetic energy is converted into nuclear energy the nuclei must overcome a strong electrostatic repulsion 1.9 × 109 K 2.9 × 109 K 4.3 × 109 K 1.2 × 1028 K The energy of a single photon of the ultraviolet laser is A. B. C. D. 7.1 × 10–19 J 1.0 × 10–27 J 7.1 × 10–28 J 1.9 × 10–40 J 21 21 35. The absorption spectrum of hyd...
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This note was uploaded on 01/14/2014 for the course PHYSICS Physics 30 taught by Professor Quinlan during the Fall '09 term at Centennial High School.

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