ch28-p037 - n (200 eV) = m p v 2 /2 in this particular...

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37. (a) By conservation of energy (using qV for the potential energy which is converted into kinetic form) the kinetic energy gained in each pass is 200 eV. (b) Multiplying the part (a) result by n = 100 gives K = n (200 eV) = 20.0 keV. (c) Combining Eq. 28-16 with the kinetic energy relation (
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Unformatted text preview: n (200 eV) = m p v 2 /2 in this particular application) leads to the expression r = m p e B 2 n (200 eV) m p . which shows that r is proportional to n . Thus, the percent increase defined in the problem in going from n = 100 to n = 101 is 101/100 1 = 0.00499 or 0.499%....
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This note was uploaded on 06/03/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.

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