For the velocity to double, the kinetic energy must become four times the original value. The KE increase is thee times the original value. Set that equal to the potential energy decrease.
3*(m/2)Vo^2 = k e^2/R
e is the electron charge and k is the Coulomb constant. m is the electron mass.
Solve for the unknown, R.
An electron is projected, with an initial speed of vi=1.10e+05 m/sec, directly towards a proton that is essentially at rest. If the electron is initially a great distance from the proton, at what distance from the proton is its speed instantaneously equal to twice its initial value? Express your answer in meters.
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