30_InstSolManual_PDF_Part9

30_InstSolManual_PDF_Part9 - 30.34. Set Up: The reaction...

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Unformatted text preview: 30.34. Set Up: The reaction energy is where and If neutral atom masses from Table 30.2 are used, the electron masses cancel. 1 u is equivalent to 931.5 MeV. If energy is liberated in the reaction. The threshold energy is where r is the sum of the radii of the reacting nuclei A and B . with Solve: (a) so 7.15 MeV is liberated. (b) 30.35. Set Up: Assume the separation of the proton and is the radius of the nucleus plus the radius of the proton. The nucleus has charge Treat the proton and nucleus as point charges when calculating the electrical potential energy of the two particles. To reach the surface of the nucleus the proton must have initial kinetic energy equal to the electrical potential energy when its center is a distance R from the center of the nucleus, where R is the sum of radius of the nucleus and the radius of the proton. Solve: Reflect: Our analysis assumes that the charge distribution with in the nucleus is spherically symmetric so out- side the nucleus its potential is the same as if all its charge were concentrated at its center. We have also assumed thatside the nucleus its potential is the same as if all its charge were concentrated at its center....
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