#$&* course Phy 122 *********************************************
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Given Solution: The change in potential is final potential - initial potential = -55 V - (135 V) = -190 V, so the change in the potential energy of the proton is -190 V * 1.6 * 10^-19 C = -190 J / C * 1.6 * 10^-19 C = -3.0 * 10^-17 J. In the absence of dissipative forces this is equal and opposite to the change in the KE of the proton; i.e., the proton would gain 3.09 * 10^-17 J of kinetic energy. Change in potential energy is equal and opposite to the work done by the field on the charge, so the field does 3.0 * 10^-17 J of work on the charge. Since the charge of the proton is equal in magnitude to that of an electron, he work in electron volts would be 180 volts * charge of 1 electron= 180 eV. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `qQuery Principles and General Physics 17.8: Potential difference required to give He nucleus 65.0 keV of KE. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: 6.5e4 voles Confidence rating: 3
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Given Solution: 65.0 keV is 65.0 * 10^3 eV, or 6.50 * 10^4 eV, of energy. The charge on a He nucleus is +2 e, where e is the charge on an electron. So assuming no dissipative forces, for every volt of potential difference, the He nuclues would gain 2 eV of kinetic energy. To gain 6.50 * 10^4 eV of energy the voltage difference would therefore be half of 6.50 * 10^4 voles, or 3.35 * 10^4 volts. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!