problem43_54

problem43_54 - d) , Po At 210 84 210 85 m m so the decay is...

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43.54: a) MeV. 41 . 5 or u, 10 81 . 5 3 He Pb Po 4 2 206 82 210 84 = × = - - - Q m m m The energy of the alpha particle is ) 210 206 ( times this, or MeV 30 . 5 (see Example 43.5) b) , 0 u 10 35 . 5 3 H Bi Po 1 1 209 83 210 84 < × - = - - - m m m so the decay is not possible. c) 0, u 10 22 . 8 3 n Po Po 209 84 210 84 < × - = - - - m m m so the decay is not possible.
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Unformatted text preview: d) , Po At 210 84 210 85 m m so the decay is not possible (see Problem (43.50)). e) , 2 Po e Bi 210 84 210 83 m m m + so the decay is not possible (see Problem (43.51))....
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This note was uploaded on 05/12/2008 for the course PHYS 262 taught by Professor Dougherty during the Spring '08 term at Lafayette.

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