Electrons in orbit around a proton: The smallest has aradius of .0529 nm. The ne

Question

Electrons in orbit around a proton: The smallest has aradius of .0529 nm. The next largest has a radius of .212nm.Suppose the electron were moved from the smallest to the nextorbit, what would be the change in the electric potentialenergy of the electron? qp= +1.60 x 10-19 C r1= .0529 nm = 5.29 x 10-11m r2= .212 nm = 2.12 x 10-10 m(1nm=10-9 m) V1= +27.2 V V2= +6.79 V Electrons in orbit around a proton: The smallest has aradius of .0529 nm. The next largest has a radius of .212nm.Suppose the electron were moved from the smallest to the nextorbit, what would be the change in the electric potentialenergy of the electron? Electrons in orbit around a proton: The smallest has aradius of .0529 nm. The next largest has a radius of .212nm.Suppose the electron were moved from the smallest to the nextorbit, what would be the change in the electric potentialenergy of the electron? qp= +1.60 x 10-19 C r1= .0529 nm = 5.29 x 10-11m r2= .212 nm = 2.12 x 10-10 m(1nm=10-9 m) V1= +27.2 V V2= +6.79 V

Explanation / Answer

Electric Potential Energy (Ue) = qV So plugging in the values for qp and the voltages forthe small and large radius we get: Ue = qp (V1 - V2)= (+1.60 x 10-19 C)(27.2 V - 6.79 V) Ue = + 3.27 x 10-18 J

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