1194_Physics ProblemsTechnical Physics

1194_Physics - Chapter 42(b 535 Solving(3 for n gives n = GM S r ME(4 Taking MS = 1.99 10 30 kg and M E = 5.98 10 24 kg r = 1 m = 1 Js we find G =

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Chapter 42 535 (b) Solving (3) for n gives nG M r M S E = = . (4) Taking M S 199 10 30 . k g , and M E 598 10 24 k g , r 1496 10 11 m , G 667 10 11 N m k g 22 , and = 1055 10 34 . Js , we find n 253 10 74 .. (c) We can use (3) to determine the radii for the orbits corresponding to the quantum numbers n and n + 1 : r n GM M n SE = 2 = and r n GM M n + = + 1 2 2 2 1 af = . Hence, the separation between these two orbits is r GM M nn GM M n =+ = + == 2 2 2 2 2 2 12 1 a f . Since n is very large, we can neglect the number 1 in the parentheses and express the separation as r GM M n ≈= × = 2 2 63 2 1 18 10 a f m . This number is much smaller than the radius of an atomic nucleus ~10 15 m ej , so the distance between quantized orbits of the Earth is too small to observe. *P42.53 (a) E eB m e ×× × ⋅⋅ F H G I K J F H G I K J = −− = 160 10 526 2 9 11 10 975 10 609 19 34 31 23 . . C 6.63 10 J s T kg Ns TCm kg m J eV
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This note was uploaded on 12/14/2011 for the course PHY 203 taught by Professor Staff during the Fall '11 term at Indiana State University .

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