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Finding the kinetic energy of a proton in megaelectron volts?
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[QUOTE="Simon Bridge, post: 4535957, member: 367532"] Approaching the nucleus charge +Ze, the proton charge +e is exchanging kinetic energy with potential energy. It has to start out with at least as much kinetic energy as the difference in potential energy between it's start and finish radii. You know an equation for potential energy of a charge at a distance r from another charge. Basically you seem to have confused eV with V. 1eV is the kinetic energy change in an electron moving through a potential difference of 1V i.e. it is a unit of energy... so you can convert directly from Joules. ... but it is best practice to do your calculations with more convenient units - like: ##\small ke^2=1.4400\text{MeV.fm}## ... since you have radius in fm and you want energy in MeV. [url]http://www.phas.ubc.ca/~mcmillan/rqpdfs/3_particle_nature_of_matter.pdf[/url] Similarly ##\small m_pc^2=938\text{MeV}## and ##\small E_K=(\gamma-1)m_pc^2 \approx \frac{1}{2}(m_pc^2)(v/c)^2## ... let's you get the speed in terms of the speed of light. You'll find these numbers much much easier to remember and use than SI units ;) [/QUOTE]
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Finding the kinetic energy of a proton in megaelectron volts?
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