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Finding the orbital velocity of the atomic electron, ionizing, particles & collisions

A hydrogen atom when in its lowest energy state consists of a proton of charge +e and an electron of charge -e and mass 9.1x10-31 kg. In the Bohr model of the atom, the electron moves around the nucleus in a circular orbit of radius 0.51x10-10 m. Determine the speed that another electron, starting very far away from the hydrogen atom, must have in order to ionize the atom during a collision. In the final state, all three particles (proton and two electrons) are considered at rest very far from each other.

(a) Construct a pictorial representation showing clearly the initial and final situations.

(b) Choose a system. Construct work-energy bar graphs for this process, showing Initial Energy + Wout = Final energy. (Hint do not ignore the circular velocity of the orbiting electron).

(c) Based on your work-energy bar graphs, write down the work-energy theorem for this problem. Use Newton's laws to find the orbital velocity of the atomic electron.

(d) Solve for the initial velocity of the second electron.

Solution Preview

See attached file.

charge on an electron = on proton = e = 1.6*10(-19) C
mass of electron (me) = 9.1*10^(-31) kg
radius of orbital motion of the electron (r) = 0.51*10^(-10) m

a.)
See attached file.

b.)
Initial energy = Potential energy of electron proton system (U) + k.E. of electron orbiting the proton in an orbit(K)
U = -9*10^9 * e^2/r (-ve sign, because elect. and proton. have opp.charges)
=> U= -9*10^9*(1.6*10^(-19))^2/(0.51*10^(-10))
=> U = 45.18 ...

Solution Summary

With careful calculations and good explanations, the problem is solved including a pictorial.

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