Unformatted text preview: − 2 K 2 E 1 + ( E 1 − E 2 ) 2 = E 2 3 . This is now straightforward to solve for K 2 and yields the result stated in the problem. (b) Setting E 3 = 0 in K 2 = 1 2 E 1 h ( E 1 − E 2 ) 2 − E 2 3 i and using the rest energy values given in Table 451 readily gives the same result for K µ as computed in Sample Problem 451....
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This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.
 Fall '08
 SPRUNGER
 Physics, Conservation Of Energy, Energy

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