JEE Advanced 2012 Physics question
A proton is fired from very far away towards a nucleus with charge Q = 120 e, where e is the electronic charge. It makes a closest approach of 10 fm to the nucleus. The de Broglie wavelength (in units of fm) of the proton at its start is: (take the proton mass, kg; J.s/C; m/F; 1 fm = m)
Answer: 7
Solution
At closest approach the initial kinetic energy is entirely converted into electrostatic potential energy: J. Momentum \sqrt{2m_pK}$ =
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