Constants: e=1.602 x 10-19 C me=9.1 x 10-31 kg co=8.854 x 10-12 F/m • 440 = 4 x 107 H/m c=√1/50440 = 3 x 108 m/s 6.64 x

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answerhappygod
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Constants: e=1.602 x 10-19 C me=9.1 x 10-31 kg co=8.854 x 10-12 F/m • 440 = 4 x 107 H/m c=√1/50440 = 3 x 108 m/s 6.64 x

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Constants E 1 602 X 10 19 C Me 9 1 X 10 31 Kg Co 8 854 X 10 12 F M 440 4 X 107 H M C 1 50440 3 X 108 M S 6 64 X 1
Constants E 1 602 X 10 19 C Me 9 1 X 10 31 Kg Co 8 854 X 10 12 F M 440 4 X 107 H M C 1 50440 3 X 108 M S 6 64 X 1 (46.18 KiB) Viewed 18 times
Constants: e=1.602 x 10-19 C me=9.1 x 10-31 kg co=8.854 x 10-12 F/m • 440 = 4 x 107 H/m c=√1/50440 = 3 x 108 m/s 6.64 x 10-27 kg) is accelerated Problem 1 (10 points): An alpha particle (Q₁ = +2e, m = towards a gold atom (Q2 = +79e) by a potential U = 10 kV. As the particle approaches the gold nucleus, it decelerates due to the Coulomb force. Use conservation of energy to determine how close the alpha particle will come to the gold atom. Problem 2 (10 points): The hydrogen atom has one proton and one electron, and the Bohr radius is ao = 5.3 x 10-11 m. Determine the orbital velocity of the electron. C₂ V = G₁₂ C3 AY
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