If an electron can orbit a proton only in orbits having angular momentum h/(2`pi), 2 * h/(2`pi), 3 * h/(2`pi), ... , n * h/(2`pi), ..., then what is the radius of the closest possible orbit? How many deBroglie wavelengths are required to span the circumference of this orbit? What are the radii of the two next-closest orbits? What is the radius of the nth-closest orbit? What it is the total energy of each of these orbits? How many deBroglie wavelengths are required to span the circumference of each orbit?
SolutionIf an electron 'orbits' a proton at distance r, then it experiences the Coulomb force
Its centripetal acceleration must therefore be
where mE is the mass of the electron.
We can therefore find from the relationship aCent = v^2 / r the velocity of the electron:
If in addition we have angular momentum mE * v * r = n * h / ( 2 `pi ) for positive integer n, we see that v = n h / ( 2 `pi mE r). Combining this with the expression v = `sqrt( k / (r * mE ) ) * qE for v, we have
Solving for r we have the follow for rn, the radius of the nth possible orbit:
Since r is proportional to n, we see that the closest orbits are for the smallest values of n. We see also that orbital radii are proportional to n^2, so the relative radii are 1, 4, 9, 16, ... times that of the closest radius.
We find the radius of the closest orbit by evaluating for n = 1:
The next two orbits would have radii
Note that the only difference between this radius and the first is the 2^2, as opposed to the 1^2, at the beginning of the calculation. We could have obtained this result more easily by multiplying the radius of the first orbit by 2^2 / 1^2 = 4:
The radius of the third orbit is most easily calculated by a similar use of proportionality:
The numerical radius of the nth orbit would be
The velocity of the electron in the nth orbit would be
Evaluating vn = 1 / n * [ 2 `pi k qE^2 / h ] for accepted values of k, qE and h we have
The velocity of the electron in the first orbit would be 2.17 * 10^6 m/s; the velocity in the second is 1 / 2 * [ 2.17 * 10^6 m/s ]; subsequent velocities are 1 / 3 * [ 2.17 * 10^6 m/s ], 1 / 4 * [ 2.17 * 10^6 m/s ], etc.
The PE of the nth orbit is found from r(n):
The KE of the nth orbit is found from vn
which differs from the expression for the PE by the lack of a negative sign and by the 2 `pi^2 instead of the 4 `pi^2--i.e., this KE is -1/2 of the PE. This proportionality between KE and PE in an orbit is by now familiar .
The numerical approximation to this energy is
The total energy is therefore the sum of these energies
The first few kinetic energies are thus
The energy required to change from one 'orbit' n1 to 'orbit' n2, using the 13.6 eV approximations to the common factor 2 `pi^2 k^2 mE qE^4 / h^2 is thus
= -1 / n1^2 * [ 13.6 eV] - { -1 / n2^2 * [ 13.6 eV] }
= ( 1 / n2^2 - 1 / n1^2 ) * [ 13.6 eV] .
The deBroglie wavelength of the nth orbit is
Comparing this to the circumference 2 `pi r(n) = 2 pi ( n^2 h^2 / ( 4 `pi^2 k mE qE^2 ) ) = n^2 h^2 / ( 2 `pi k mE qE^2), we see that the two agree except for the factor n^2 in the circumference, as opposed to n in the expression for the wavelength. Thus we have
so that n deBroglie wavelengths of the electron fit into the nth orbit.
Working this out numerically we see that `lambda(n) = n [ h^2 / ( 2 `pi k mE qE^2) ] = n * 3.32 Angstroms. So the first three deBroglie wavelengths are
The corresponding circumferences are
The ratios of circumference to wavelength are thus
We thus visualize standing waves with 1, 2 and 3 wavelengths 'wrapped around' the respective circular orbits, forming a continuous cyclical wave.
General SolutionThis is a synopsis of the above solution in symbolic terms only:
If an electron 'orbits' a proton at distance r, then it experiences the Coulomb force
Its centripetal acceleration must therefore be
where mE is the mass of the electron.
We can therefore find from the relationship aCent = v^2 / r the velocity of the electron:
If in addition we have angular momentum mE * v * r = n * h / ( 2 `pi ) for positive integer n, we see that v = n h / ( 2 `pi mE r). Combining this with the expression v = `sqrt( k / (r * mE ) ) * qE for v, we have
Solving for r we have the follow for rn, the radius of the nth possible orbit:
The corresponding velocity is found by substituting this expression for r into v = n h / ( 2 `pi mE r ):
The momentum of the electron at this velocity is
The deBroglie wavelength `lambda = h / p is thus
The number of wavelengths in the circumference will therefore be
Thus n deBroglie wavelengths fit into one 'orbit'.
The corresponding KE and PE are
and
Note that again PE = -2 * KE.
Total energy is
The energy required to change from one 'orbit' n1 to 'orbit' n2 is thus