42. (a) The potential isV(r)=14πε0er=³8.99×109 N·m2C2´(1.60×10−19C)5.29×10−11m=27.2V.(b) The potential energy isU=−eV(r−27.2eV.(c) Sincemev2/r=−e2/40r2,K=12mv2=−12µe240r¶=−12V(r27.2=13.6eV.(d) The energy required is
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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.