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What is the magnitude p of the electron's momentum, in the unit MeV/c? (Note that c is the symbol for the speed of light and not itself a unit.) |
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Answer» <P> Solution :KEY IDEAWe can find p from the total energy E and the mass energy `mc^(2)` , `E^(2)=(pc)^(2)+(mc^(2))^(2)`. CALCULATIONS: Solving for pc gives US `pc=sqrt(E^(2)-(mc^(2))^(2))` `=sqrt((3.04MeV)^(2)-(0.511MeV)^(2))=3.00MeV`. Finally, dividing both sides by c we find `p=3.00MeV//c`. |
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