1208_Physics ProblemsTechnical Physics

# 1208_Physics ProblemsTechnical Physics - Chapter 43(c af F...

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Chapter 43 549 (c) Fr dU dr r r a f =− = F H G I K J F H G I K J L N M M O Q P P 4 12 6 13 7 σ σσ To find the maximum force we calculate dF dr r r = F H G I K J + F H G I K J L N M M O Q P P = 4 156 42 0 2 14 8 when r rupture = F H G I K J 42 156 16 F max . . .. . . = F H G I K J F H G I K J L N M M O Q P P × 4465 0272 12 42 156 6 42 156 41 0 41 0 16 10 655 13 6 7 6 19 9 eV nm eV nm Nm 10 m nN af Therefore the applied force required to rupture the molecule is + . n N away from the center. (d) Ur s rs E r E aa 0 0 12 0 6 0 0 12 0 6 44 2 2 +=∈ + F H G I K J + F H G I K J L N M M O Q P P + F H G I K J + F H G I K J L N M M O Q P P + bg =∈ + F H G I K J −+ F H G I K J L N M M O Q P P + + F H G I K J −−+ F H G I K J L N M M O Q P P + = −∈+ ∈ −∈ +∈− ∈ + + =−∈+ + F H G I K J +∈+ +≅ −− 4 1 4 1 1 2 1 4 1 4 11 2 7 8 1 2 2 1 12 78 2 12 42 03 6 0 12 0 6 0 2 0 2 0 2 0 2 0 2 0 2 0 2 0 2 0 2 0 2 0 s r s r E s r s r s r s r E s r s r s r s r E E s r s r s Ur a a a a …… 0 2 1 2 + ks where k
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