Under the influence of the electric force of attraction, the electron in a hydrogen atom orbits around the proton on a circle of radius 5.3 x 10^-11 m. What is the orbital speed? What is the orbital period?
The centripetal force m V^2/r must equal the Coulomb attraction force ke^2/r^2.
Solve for v, the orbital speed.
The period will be
P = 2 pi r/ V
To find the orbital speed and orbital period of the electron in a hydrogen atom, we can use the concepts of centripetal force and centripetal acceleration.
The electric force of attraction provides the centripetal force required to keep the electron in orbit. This force is given by:
F = (mv^2) / r
Where:
F = Force (electric force of attraction)
m = Mass of the electron
v = Orbital speed of the electron
r = Radius of the circular orbit
We know that the electric force of attraction is equal to the gravitational force between the electron and the proton (since we are dealing with hydrogen atom):
F = k(q1*q2)/r^2
Where:
k = Coulomb's constant
q1, q2 = Charges of the electron and proton (equal in magnitude but opposite in sign)
r = Radius of the circular orbit
Since the charges of the electron and proton are equal in magnitude but opposite in sign, we can write:
F = k(e^2)/r^2
Since the electric force of attraction is equal to the centripetal force, we can equate the two equations:
k(e^2)/r^2 = (mv^2)/r
We can rearrange this equation to solve for the orbital speed:
v^2 = (k(e^2))/r
v = sqrt((k(e^2))/r)
Substituting the given values:
k = 8.99 x 10^9 Nm^2/C^2 (Coulomb's constant)
e = 1.6 x 10^-19 C (charge of the electron)
r = 5.3 x 10^-11 m (radius of the circular orbit)
v = sqrt(((8.99 x 10^9 Nm^2/C^2)(1.6 x 10^-19 C)^2)/(5.3 x 10^-11 m))
Calculating this will give us the value for the orbital speed of the electron in m/s.
To find the orbital period, we can use the formula:
T = (2πr) / v
T = (2π(5.3 x 10^-11 m)) / v
Calculating this will give us the value for the orbital period of the electron in seconds.