In a hydrogen atom (i.e., one stationary proton and one orbiting electron), the angular velocity of the electron is 10 x 10^6 ra
1 answer:
Assuming the centripetal force is provided by the Coulomb attraction between the electron and the proton we have that,
Here,
k = Coulomb's Constant
R = Distance
e = Electron charge
And by the centripetal force,
= Mass of electron
R = Radius
= Angular velocity
Equation both expression,
Replacing,
Therefore the radius of the electron orbit is
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Answer:
T = 1.12 10⁻⁷ s
Explanation:
This exercise must be solved in parts. Let's start looking for the electric field in the axis of the ring.
All the charge dq is at a distance r
dE = k dq / r²
Due to the symmetry of the ring, the field perpendicular to the axis is canceled, leaving only the field in the direction of the axis, if we use trigonometry
cos θ =
dEₓ = dE cos θ
cos θ = x / r
substituting
dEₓ =
DEₓ = k dq x / r³
let's use the Pythagorean theorem to find the distance r
r² = x² + a²
where a is the radius of the ring
we substitute
dEₓ =
we integrate
∫ dEₓ =k \frac{x}{(x^2 + a^2 ) ^{3/2} } ∫ dq
Eₓ =
In the exercise indicate that the electron is very central to the center of the ring
x << a
Eₓ =
if we expand in a series
we keep the first term if x<<a
Eₓ =
the force is
F = q E
F =
this is a restoring force proportional to the displacement so the movement is simple harmonic,
F = m a
the solution is of type
x = A cos (wt + Ф)
with angular velocity
w² =
angular velocity and period are related
w = 2π/ T
we substitute
4π² / T² = \frac{keQ}{m a^3}
T = 2π
let's calculate
T = 2π
T = 2π pi
T = 2π 17.9 10⁻⁹ s
T = 1.12 10⁻⁷ s