An electron is projected, with an initial speed of vi=2.06e+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.
To find the distance from the proton at which the electron's speed is instantaneously equal to twice its initial value, we can use the conservation of mechanical energy.
The initial speed of the electron, vi, is given as 2.06e+05 m/s.
Let's assume that the distance between the electron and proton at this point is x.
The final speed of the electron is twice its initial value, which is 2 * vi = 2 * 2.06e+05 = 4.12e+05 m/s.
Using the conservation of mechanical energy, we can equate the initial kinetic energy of the electron to the final kinetic energy of the electron:
(1/2) * me * vi^2 = (1/2) * me * vf^2
Where me is the mass of the electron. The mass of the proton is much larger than the electron and can be considered at rest.
Since the masses of the electron and proton are the same, we can cancel them out:
vi^2 = vf^2
Substituting the given values:
(2.06e+05)^2 = (4.12e+05)^2
Solving this equation, we find the square of the distance, x^2, is:
x^2 = (2.06e+05)^2 - (4.12e+05)^2
x^2 = 4.2436e+09 - 1.6974e+10
x^2 = -1.2731e+10
Since distance cannot be negative, we are unable to find a real solution for x. This means there is no distance from the proton at which the electron's speed is instantaneously equal to twice its initial value.