Question

of radius 1.71 x 10 m around the proton. Assume the orbital angular momentum of the electronis In the Bohr model of the hydrogen atom, the electron in the no 18 level moves in equal to 18/2. (a) Calculate the orbital speed of the electron ms (b) Calculate the kinetic energy of the electron (c) Calculate the angular frequency of the electron's motion rad/s

Answer 1

(a) orbital speed can be found by two ways

v = nh / 2$\pi$mr

where m is mass of electron, n is orbit number

and

the other more simple formula is

v = 2.18e6 * z / n

where z is atomic number of atom ( note that this formula is valid for atoms with single electron)

so,

v = 18 * 6.6e-34 / 2$\pi$ * 9.11e-31 * 1.71e-8

v = 1.21e5 m/s

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(b)

energy can be found as

E = - 13.6 / n² ( in eV)

or

E = - 2.18e-18 / n² ( in J)

so,

putting n = 18

E = 0.041975 eV

or

E = 6.72e-21 J

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(c)

angular frequency, w = v / r

w = 1.21e5 / 1.71e-8

w = 7.1e12 rad/sec