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.
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.
To find the distance from the proton where the speed of the electron is twice its initial value, we need to use conservation of mechanical energy.
The initial mechanical energy (Ei) of the electron is given by the sum of its kinetic energy (KEi) and its electric potential energy (PEi):
Ei = KEi + PEi
Since the proton is essentially at rest, the electric potential energy is zero. Therefore, the initial mechanical energy of the electron is only kinetic energy:
Ei = KEi
The kinetic energy (KE) of the electron is given by the equation:
KE = (1/2)mv^2
where m is the mass of the electron and v is its velocity.
To find the distance where the speed of the electron is twice its initial value, we equate the kinetic energy at the new velocity (2v) to the initial kinetic energy (v):
(1/2)me(2v)^2 = (1/2)me(vi)^2
Cancelling out the constants and simplifying:
(2v)^2 = (vi)^2
4v^2 = (1.10e+05 m/sec)^2
Taking the square root on both sides:
2v = 1.10e+05 m/sec
Solving for v:
v = 0.55e+05 m/sec
Now, we can use kinematic equations to find the distance (d) at which the speed of the electron is twice its initial value.
The equation for velocity (v) in terms of initial velocity (vi), acceleration (a), and distance (d) is:
v^2 = vi^2 + 2ad
Substituting the known values:
(2v)^2 = (vi)^2 + 2ad
(1.10e+05 m/sec)^2 = (1.10e+05 m/sec)^2 + 2a(d)
simplifying:
1.10e+05 m/sec = 2ad
Now, we know that the initial velocity (vi) is 1.10e+05 m/sec and the final velocity (v) is 2v, which we found to be 0.55e+05 m/s. Therefore, substituting these values into the equation:
(0.55e+05 m/sec)^2 = (1.10e+05 m/sec)^2 + 2a(d)
Simplifying:
0.30e+05 m/sec = 2ad
Finally, solving for distance (d):
d = (0.30e+05 m/sec) / (2a)
where a is the acceleration of the electron due to the electric attraction towards the proton.
Substituting the value of the electron charge (e) and the permittivity of free space (ε0) into the equation:
d = (0.30e+05 m/sec) / (2 * (1.602e-19 C)^2 / (9.0e+09 N m^2/C^2))
Evaluating the expression:
d = 6.76e-11 m
Therefore, the distance from the proton where the speed of the electron is instantaneously equal to twice its initial value is approximately 6.76e-11 meters.