The ground state orbit in the Bohr atom model of the hydrogen atom has a radius

Question

The ground state orbit in the Bohr atom model of the hydrogen atom has a radius of 5.2912 x 10^-11 m (about 0.53 degree A). Although the Bohr model is no longer considered an adequate model of the atom, the Bohr radius often shows up as a useful constant in quantum mechanics. (a) Using your derivations from the second problem in problem set 2 (the Rutherford atom), determine the kinetic energy of an electron in this orbit. (b) What is the linear momentum of the electron? (c) Debroglie proposed (in his PhD dissertation) that the electron existed in this orbit because of its wavelike properties. Determine the DeBroglie wavelength of the electron in the ground state of the hydrogen atom if it circles the nucleus at the Bohr radius.

Explanation / Answer

(a)

The force of attraction acting on the electron due to the proton in the Bohr's atom is :

Fe = k*e2/r2

Here, k = 9*109 N m2/C2

e = Charge on electron and proton = 1.6*10-19 C

r = Bohr's radius = 5.29*10-11 m

Centripetal force, Fr = mv2/r

Here, m = mass of electron , v = velocity of electron in Bohr's orbit

For equilibrium,

Fe = Fr

Thus, mv2/r = ke2/r2

Thus, kinetic energy, KE = 0.5mv2 = 0.5*k*e2/r = 0.5*9*109*(1.6*10-19)/(5.29*10-11) = 2.177*10-18 J

(b)

Velocity of electron in this orbit, v = 2.19*106 m/s

Mass of electron, m = 9.1*10-31 kg

So, linear momentum of electron, p = mv = 9.1*10-31*2.19*106 = 19.929*10-25 kg m/s

(c)

The De-Broglie wavelength of a particle is given by :

l = h/p , where

h = Planck's constant = 6.626*10-34 m2 kg/s

p = linear momentum of particle

So, De-Broglie wavelength, l = (6.626*10-34)/(19.929*10-25) = 33.2*10-11 m

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