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quiz4_key_c-b-a_f08

# quiz4_key_c-b-a_f08 - CHEM 111A TIME ALLOWED 15 MINUTES...

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CHEM 111A NAME: _______KEY__ ___________ TIME ALLOWED: 15 MINUTES ID# ____________________________ Quiz 4.c 10/2/08 All Quizzes are closed book and closed notes. *** Show your work/reasoning *** General Information: mv h c =  p = mv 4 h p x E K = ½ mv 2 E = h h = 6.626 x 10 -34 Js m electron = 9.11 x 10 -31 kg m proton = 1.67 x 10 -27 kg N A = 6.022 x 10 23 1.6022 x 10 -19 J = 1 eV 10 10 Å = 1 m 10 9 nm = 1 m 10 12 pm = 1 m c = 2.998 x 10 8 m/s 1. a) In the ground state, the hydrogen atom has a radius on the order of 0.05 nm and a kinetic energy of 13.6 eV Assuming that we know the position of an electron to an accuracy of 1% of the hydrogen radius, calculate the uncertainty in the velocity of the electron (in m/s). Please circle your answer. 4 h p x m 10 x 0 . 5 nm 10 m 1 ) nm 05 . 0 )( 01 . 0 ( x 13 9 s / m kg 10 x 05 . 1 J 1 s m kg 1 ) m 10 x 0 . 5 )( 4 ( Js 10 x 626 . 6 ) x ( 4 h p 22 1 2 13 34 v m mv p and so m p v s / m 10 x 16 . 1 kg 10 x 11 . 9 s m kg 10 x 05 . 1 v 8 31 1 22 b) Is the electron’s uncertainty in velocity large or small? Justify your answer with a calculation or comparison. With such a small uncertainty in the electron’s position, there is a minimum uncertainty of 1.16 x 10 8 m/s in the electron’s velocity. At this level of uncertainty, we have virtually no idea of the velocity of the electron. MUST state that his is a very large number, of the same magnitude as the speed of light (3 x 10 8 m/s). OR, calculate the

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quiz4_key_c-b-a_f08 - CHEM 111A TIME ALLOWED 15 MINUTES...

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