Answer Key
Question and Answer - Nuclear Physics #2
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University
Princeton University Course
PHY 103 | General Physics IPages
2
Academic year
2023
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abdulsalam23745
Views
0
cues. What Should be the minimum K.E of the electron to probe the size of 4-0 40 20Ca nucleus ? 20Ca sol Rca = RoA 13 = 1.2(40) 1/3 = 4 -fm 40 In order to prope 20 Ca nucleus the de-broglie wavelength 1= h of incident particle should P to be equal to 8fm (diameter) of 20 Ca nucleus. Total energy Mc2 Moc2 E = = E= Moc2 +K , C2 M&C4 + meet E = p2c2 + 11 1 = (hc)2 + m2 12
MeC2 = 0.511 Mev tic = 197.3 Mek fm, he = 197.3.22x3.14 = 1239 Mev-fm E = (1239)2 (mev-fm) 8 fm2 + (0.511)2 = 23986.526 Mev = 154.87 MeV K = E-Mec2 = 154-87-0.511 = 154.36 MeV
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Question and Answer - Nuclear Physics #2
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