Find the magnitude of the electrical force and the
gravitational force between the electron and the proton in the
atom. The average distance between the two particles is 0.53 A
Fe=qeqp4πε0R2=(1.6×10−19)24π(8.85×10−12)(0.53×10−10)2=2.56×10−38(1.112×10−10)(0.2809×10−20)=2.56×10−383.1236×10−31=8.2×10−8 NF_e = \frac{q_eq_p}{4πε_0R^2} \\ = \frac{(1.6 \times 10^{-19})^2}{4π(8.85 \times 10^{-12})(0.53 \times 10^{-10})^2} \\ = \frac{2.56 \times 10^{-38}}{(1.112 \times 10^{-10})(0.2809 \times 10^{-20})} \\ = \frac{2.56 \times 10^{-38}}{3.1236 \times 10^{-31}} \\ = 8.2 \times 10^{-8} \;NFe=4πε0R2qeqp=4π(8.85×10−12)(0.53×10−10)2(1.6×10−19)2=(1.112×10−10)(0.2809×10−20)2.56×10−38=3.1236×10−312.56×10−38=8.2×10−8N
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