In the Bohr model of the hydrogen atom, the atom is held together by the electro

Question

In the Bohr model of the hydrogen atom, the atom is held together by the electrostatic attraction between the proton and electron, and the electron moves in a circular orbit around the stationary proton. Find the radius of the electron’s orbit and the speed of the electron when the atom is in the ground state. The ground-state energy of the hydrogen atom is -13.6 eV.

Hint: Recall that an object moving with speed v along a circular path of radius r experiences a centripetal acceleration of magnitude ac = v2/r. Begin by applying Newton’s second law to the electron and show that the K = -12 U, where K is the kinetic energy of the electron and U is its potential energy. (You can do this without plugging in any numbers.)

Possibly useful information:
mass of the electron = 9.1110-31 kg mass of the proton = 1.67310-27 kg 1 eV = 1.610-19 J

Explanation / Answer

centripetal force = electrostatic force

m*v^2/r = Zk*e^2/r^2

v = sqrt(Z*k*e^2/(m*r))

and apply m*v*r = nh/(2*pi)

for ground state n=1 and fore hydrogen Z=1

m* sqrt(Z*k*e^2/(m*r))*r =h/(2*pi)

r = h^2/(4*pi^2*k*e^2*m) = 5.29*10^-11 m

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