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

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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 Summary

The orbital velocity of the atomic electron is examined. The expert construct a pictorial representation showing the initial and final situations. With careful calculations and good explanations, the problem is solved including a pictorial.

##### 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 ...

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###### Education
• MSc , Pune University, India
• PhD (IP), Pune University, India
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