Suppose an electron could orbit a proton at a constant distance of 1.14 Angstroms, with the the centripetal force coming from the Coulomb attraction between proton and electron.

• What would be its orbital velocity, orbital KE, electrostatic PE relative to infinite separation, and total energy?
• What would be its deBroglie wavelength at the orbital velocity?
• How many deBroglie wavelengths are required to span the circumference of this orbit?   What does this say about the possibility of actually achieving this orbit?

An electron at distance 1.14 Angstroms = 1.14 * 10^-10 m from a proton will experience a force of magnitude | F | = k | q1 q2 | / r^2, so

• force of attraction between electron and proton = 9 * 10^9 N m^2 / C^2 * (1.6 * 10^-19 C) ^ 2 / ( 1.14 * 10^-10 m ) ^ 2 = 17.72853 * 10^-19 N

The resulting centripetal acceleration of the electron is therefore

• centripetal acceleration = F / m = 17.72853 N / ( 9.11 * 10^-31 kg) = 1.946052 * 10^22 m/s^2.

Since this acceleration is aCent = v^2 / r, we have v = `sqrt( aCent * r ) so

• velocity of electron = `sqrt ( ( 1.946052 * 10^22 m/s^2) * ( 1.14 * 10^-10 m) ) = 1.489463 * 10^6 m/s.

This electron therefore has momentum p = m v. Thus

• momentum = p = (9.11 * 10^-31 kg) * 1.489463 * 10^6 m/s = 13.56901 * 10^-25 kg m/s.

The kinetic energy of the electron is .5 m v^2, or

• KE = .5 ( 9.11 * 10^-31 kg) * ( 1.489463 * 10^6 m/s) ^ 2 = 10.10526 * 10^-19 J.

The potential energy of the system is PE = k q1 q2 / r so

• potential energy = 9 * 10^9 N m^2 / C^2 * (1.6 * 10^-19 C) ( -1.6 * 10^-19 C) / ( 1.14 * 10^-10 m ) ^ 2 = -20.21053 * 10^-19 J.

The total energy is the sum of the KE and the PE:

• total energy = KE + PE = ( 10.10526 * 10^-19 J) + ( PE * 10^-19 J) = -10.10526*10^-19 J.

The deBroglie wavelength of the electron is `lambda = h / p. Therefore

• wavelength = 6.63 * 10^-34 J s / ( 13.56901 * 10^-25 kg m/s ) = 4.893506 * 10^-10 m.

Since the circumference of the orbit is 2 `pi r, we see that the number of wavelengths in the circumference is

• # of wavelengths fitting around circumference = 2 `pi * 1.14 * 10^-10 m / ( 4.893506 * 10^-10 m) = 1.463746.

Unless this number is a whole number of wavelengths, the 'electron wave' cannot fit the 'orbit' as a standing wave, since its 'ends' will not be in phase.  So this 'orbit' cannot actually occur--it would destructively interfere with itself.

As will be seen in the next problem, we will have a whole number of wavelengths in the orbit if, and only if, the angular momentum of the electron in its orbit is an integer multiple of h / ( 2 `pi).

If an electron 'orbits' a proton at distance r, then it experiences the Coulomb force

• force attracting electron to proton: F = k * qE^2 / r^2.

Its centripetal acceleration must therefore be

• centripetal acceleration: aCent = F / mE,

where mE is the mass of the electron.

We can therefore find from the relationship aCent = v^2 / r the velocity of the electron:

• electron velocity: v = `sqrt(aCent * r) = `sqrt ( F * r / mE ) = `sqrt ( ( k * qE^2 / r^2 ) * r / mE ) = `sqrt( k / ( r * mE ) ) * qE

This gives the electron a momentum of

• electron momentum: p = mE v = mE * qE * `sqrt( k / (r * mE) )

From the electron velocity we find the electron kinetic energy to be

• electron KE = .5 mE v^2 = .5 mE * [ `sqrt( k / ( r * mE ) ) * qE ] ^2 = .5 k qE^2 / r.

The potential energy of the system is found by summing the work done over small increments of distance in bringing the charges to the given proximity from infinite separation:

• potential energy: PE = -k * qE^2 / r

The total energy will be the sum of the kinetic and potential energies:

• total energy =-.5 k * qE / r.

The deBroglie wavelength `lambda = h / p is thus

• wavelength = h / p

The number of wavelengths in the circumference will therefore be

• # of wavelengths = 2 `pi r / [ h / ( mE v ) ] = ( 2 `pi / h ) * mE v r.

This will be positive integer if, and only if for some positive integer n

• condition for whole number of wavelengths: ( 2 `pi / h ) * ( mE v r ) = n,

which is the same as

• condition for whole number of wavelengths: mE v r = n h / ( 2 `pi ).

Since m v r is the angular momentum of the electron in its 'orbit', we have

• angular momentum = n h / ( 2 `pi ).