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E) 6.6 x 10 Q4. Two protons are released from rest when they are 1.00 nm apart. What will be their maximum speed? A) 1.17 x 1
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Answer #1


Here, the protons will have the maximum velocity, when they have the maximum attainable kinetic energy i.e. when all the potential energy has converted to kinetic energy.

Therefore

Uinitial = KEfinal

Uinitial=Electrostatic potential energy = kq1q2/r

Final kinetic energy of both the protons = 1/2(mv2)+ 1/2(mv2)

=> Uinitial = (9 x 109)(1.6 x 10-19)2/(1 x 10-9) = 23.04 x 10-20

=>mv2= 23.04 x 10-38

=> v = \sqrt{}(23.04 x 10-20)/(1.67x10-27) = 1.1745 x 104 m/s

option A is correct

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