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course Phy 202
8/3 at 12:30
ProblemIf 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?
Solution
R1= 1^2(6.62x10^-23)^2/((4pi^2x(9x10^9)x(9.11x10^-31)x(1.6x10^-19^2)
=0.529 angstroms
R2=2^2x(6.62x10^-23)^2/((4pi^2x(9x10^9)x(9.11x10^-31)x(1.6x10^-19^2)
=2.12 angstroms
I dont really get where to go exce
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.
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
n h / ( 2 `pi mE r) = `sqrt( k / (r * mE ) ) * qE.
Solving for r we have the follow for rn, the radius of the nth possible orbit:
r(n) = n^2 h^2 / ( 4 `pi^2 k mE qE^2 ).
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:
r(1) = 1^2 * (6.62 * 10^-23 J * s )^2 / [ 4 `pi^2 * ( 9 * 10^9 N m^2 / C^2 ) * (9.11 * 10^-31 kg) * ( 1.6 * 10^-19 C ) ^ 2 ] = .529 * 10^-10 m = .529 Angstroms.
The next two orbits would have radii
r(2) = 2^2 * (6.62 * 10^-23 J * s )^2 / [ 4 `pi^2 * ( 9 * 10^9 N m^2 / C^2 ) * (9.11 * 10^-31 kg) * ( 1.6 * 10^-19 C ) ^ 2 ] = .529 * 10^-10 m = 2.12 Angstroms.
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:
r(2) = ( 2^2 / 1^2 ) * r1 = 4 * .529 Angstroms = 2.12 Angstroms.
The radius of the third orbit is most easily calculated by a similar use of proportionality:
r(3) = ( 3^2 / 1^2 ) * r1 = 9 * .529 Angstroms = 4.76 Angstroms.
The numerical radius of the nth orbit would be
r(n) = n^2 * (6.62 * 10^-23 J * s )^2 / [ 4 `pi^2 * ( 9 * 10^9 N m^2 / C^2 ) * (9.11 * 10^-31 kg) * ( 1.6 * 10^-19 C ) ^ 2 ] = .529 * 10^-10 m = n^2 * .529 Angstroms.
The velocity of the electron in the nth orbit would be
vn = `sqrt( k / ( r(n) * mE ) ) * qE = `sqrt( k / [ n^2 h^2 mE / ( 4 `pi^2 k mE qE^2 ) ] ) * qE = 1 / n * [ 2 `pi k qE^2 / h ].
Evaluating vn = 1 / n * [ 2 `pi k qE^2 / h ] for accepted values of k, qE and h we have
vn = 1/n * [ 2.17 * 10^6 m/s].
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):
PE of nth orbit = - k qE^2 / r(n) = -k qE^2 / (n^2 h^2 / ( 4 `pi^2 k mE qE^2 ) ) = -1 / n^2 * [4 `pi^2 k^2 mE qE^4 / h^2 ] = -1/ n^2 * [ 43.5 * 10^-19 J ] = -1/n^2 * [-27.2 eV].
The KE of the nth orbit is found from vn
KE of nth orbit = .5 mE vn^2 = .5 mE ( 1 / n * [ 2 `pi k qE^2 / h ] ) ^ 2 = 1/n^2 * [ 2 `pi^2 k^2 mE qE^4 / h^2 ]
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
KE of nth orbit = 1 /n^2 * [ 13.6 eV ].
The total energy is therefore the sum of these energies
En = PEn + KEn = ... = - 1/n^2 * [ 2 `pi^2 k^2 mE qE^4 / h^2 ], or approximately -1/n^2 * 13.6 eV.
The first few kinetic energies are thus
E1 = -1 / 1^2 * 13.6 eV = -13.6 eV
E2 = -1 / 2^2 * 13.6 eV = -3.4 eV
E3 = -1 / 3^2 * 13.6 eV = -1.5 eV
E4 = -1 / 4^2 * 13.6 eV = -.85 eV.
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
energy difference = total energy of orbit n2 - total energy of orbit n1
= -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
`lambda(n) = h / p(n) = h / ( m * vn ) = h / (mE * 1 / n * [ 2 `pi k qE^2 / h ] ) = n [ h^2 / ( 2 `pi k mE qE^2) ].
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
circumference of nth orbit / wavelength of nth orbit = n^2 / n = n,
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
`lambda(1) = 1 * 3.32 Angstroms = 3.32 Angstroms,
`lambda(2) = 2 * 3.32 Angstroms = 6.64 Angstroms,
`lambda(3) = 3 * 3.32 Angstroms = 9.95 Angstroms.
The corresponding circumferences are
circumference of 1st orbit = 2 `pi r = 2 `pi * .529 Angstroms = 3.32 Angstroms
circumference of 2d orbit = 2 `pi r = 2 `pi * 2.12 Angstroms = 13.3 Angstroms
circumference of 3d orbit = 2 `pi r = 2 `pi * 4.77 Angstroms = 29.8 Angstroms.
The ratios of circumference to wavelength are thus
ratio for 1st orbit: 3.32 Angstroms / 3.32 Angstroms = 1
ratio for 2d orbit: 13.3 Angstroms / 6.64 Angstroms = 2
ratio for 3d orbit: 29.8 Angstroms / 9.95 Angstroms = 3.
We thus visualize standing waves with 1, 2 and 3 wavelengths 'wrapped around' the respective circular orbits, forming a continuous cyclical wave.
General Solution
This 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
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.
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
n h / ( 2 `pi mE r) = `sqrt( k / (r * mE ) ) * qE.
Solving for r we have the follow for rn, the radius of the nth possible orbit:
r(n) = n^2 h^2 / ( 4 `pi^2 k mE qE^2 ).
The corresponding velocity is found by substituting this expression for r into v = n h / ( 2 `pi mE r ):
v = n h / ( 2 `pi mE [ n^2 h^2 / ( 4 `pi^2 k mE qE^2 ) ] = 2 `pi k qE^2 / ( n h ).
The momentum of the electron at this velocity is
p = mE * v = mE * 2 `pi k qE^2 / ( n h ) = 2 `pi k mE qE^2 / ( n h ).
The deBroglie wavelength `lambda = h / p is thus
wavelength = h / p = h / [ 2 `pi k mE qE^2 / ( n h ) ] = n h^2 / ( 2 `pi k mE qE^2 ).
The number of wavelengths in the circumference will therefore be
# of wavelengths = 2 `pi r / wavelength = 2 `pi [ n^2 h^2 / ( 4 `pi^2 k mE qE^2 ) ] / [n h^2 / ( 2 `pi k mE qE^2 ) ] = n.
Thus n deBroglie wavelengths fit into one 'orbit'.
The corresponding KE and PE are
KE = .5 mE v^2 = .5 mE [ 2 `pi k qE^2 / ( n h ) ] ^ 2 = 1/n^2 * [ 2 `pi^2 k^2 mE qE^4 / h^2 ]
and
PE = -k qE^2 / r = -k qE^2 / [ n^2 h^2 / ( 4 `pi^2 k mE qE^2 ) ] = -1/n^2 * [ 4 `pi^2 k^2 mE qE^4 / h^2 ].
Note that again PE = -2 * KE.
Total energy is
total energy = KE + PE = -1/n^2 * [ 2 `pi^2 k^2 mE qE^4 / h^2 ] .
The energy required to change from one 'orbit' n1 to 'orbit' n2 is thus
energy difference = total energy of orbit n2 - total energy of orbit n1 = -1 / n1^2 * [ 2 `pi^2 k^2 mE qE^4 / h^2 ] - { -1 / n2^2 * [ 2 `pi^2 k^2 mE qE^4 / h^2 ] } = ( 1 / n2^2 - 1 / n1^2 ) * [ 2 `pi^2 k^2 mE qE^4 / h^2 ] .
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Problem
How much energy will an electron in orbit about a hydrogen atom lose in a transition from orbit # 2 to orbit # 6, where orbits are counted from the closest outward? If this lost energy is carried away by a photon, what will be its wavelength?
-(1/2-1/6)x13.6= 7.25x10^-19
Frequency= 7.25x10^-19/6.62x10^-34=1.10x10^15
Wavelength= 3x10^8/1.10x10^15=273.9nm
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Solution
As derived in a preceding problem we see that the difference in orbital energies from orbital n1 to orbital n2 is
( 1 / n1 - 1 / n2 ) * 13.6 eV.
Thus when n2 < n1, corresponding to a transition to a 'higher' orbit of greater radius , 1/n1 > 1/n2 so the energy difference is positive--we must add energy to the system in order to accomplish the change. Conversely when n2 > n1, we are moving to a 'lower' orbit of lesser radius and energy must be removed from the system as the change occurs.
In the present case n1 < n2 so energy must be removed from the system. The mechanism by which energy is removed is the creation of a photon with the required energy.
The energy difference in the present situation is
( 1 / n1 - 1 / n2 ) * 13.6 eV = ( 1 / 2 - 1 / 6 ) * 13.6 eV = 4.533333 eV = 7.253334 * 10^-19 J.
The energy in electron volts is easy to relate to -- just imagine electron being accelerated across the plates of a capacitor which you have charged with the hand-held generator to the appropriate number of volts.
To calculate the wavelength of the photon we prefer to use the transition energy in Joules, since this unit is compatible with the familiar units of Plank's Constant.
Since E = h f, the frequency of the photon must be
f = E / h = ( 7.253334 * 10^-19 J) ( / 6.62 * 10^-34 J s) = 1.09567 * 10^15 Hz.
The corresponding wavelength is `lambda = c / f = (3 * 10^8 m/s) / ( 1.09567 * 10^15 Hz ) = 273.8051 nm.
Generalized Solution
In general the transition energy is
E(n1, n2) = ( 1 / n1 - 1 / n2 ) * [ 2 `pi^2 k^2 mE qE^4 / h^2 ].
The corresponding frequency is
f = E(n1, n2) / h = ( 1 / n1 - 1 / n2 ) * [ 2 `pi^2 k^2 mE qE^4 / h^3 ]
and the wavelength is
`lambda = c / f = c / { ( 1 / n1 - 1 / n2 ) * [ 2 `pi^2 k^2 mE qE^4 / h^2 ] } = ( n1 n2 / (n1 + n2) ) * c / [ 2 `pi^2 k^2 mE qE^4 / h^3 ].
The last expression is numerically approximated as
numerical approximation: `lambda = ( n1 n2 / (n1 + n2) ) * 91.2 nm.
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roblem
At what Kelvin temperature will the thermal energy of electrons be greater than the energy binding an n = 3 electron to the proton in a hydrogen atom?
Solution
Energy= 1/3^2x-13.6= 1.51x(1.6x10^-19)=-2.42x10^-19
Temperature= 2/3x2.42x10^-19/1.38x10^-23
=11.68x10^3 Kelvin
The energy of the n = 3 electron is 1 / 3^2 * -13.6 eV = - 1.511111 eV = - 1.511111 eV * ( 1.6 * 10^-19 J / eV) = - 2.417778 * 10^-19 J.
It therefore requires 2.417778 * 10^-19 J of added energy to separate this electron from the atom. Thus the electron is 'bound' to the atom, requiring this much energy to separate it.
At temperature T the average thermal energy of particles is 3/2 k T, where k = 1.38 * 10^-23 J / ( particle Kelvin ) is the Boltzmann gas constant, equal to R / Navagodro.
Therefore if the average thermal energy of particles is equal to the energy we have
3/2 k T = energy,
so
T = energy / ( 3/2 k ) = (2/3) * energy / k =
[ 2/3 * 2.417778 * 10^-19 J / particle ] / [1.38 * 10^-23 J / (particle Kelvin ) ]
= 11.68009 * 10^3 Kelvin.
The result is that above this temperature the n = 3 electrons of a hydrogen atom will become separated from the atom.
Above the n=1 temperature of 11.68009 * 10^4 K all the electrons in hydrogen atoms will tend to be separated from the protons and we will have an electron-proton plasma, with a number of unique physical properties. Plasmas constitute a fourth state of matter, in addition to the liquid, solid and gaseous states.
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Problem
If in an atomic explosion a sample of .003% of the total 3 kg of fissionable material is converted to energy, how much energy is released?
Solution
Mass defect= 0.00003x3= 0.00009 kg
Energy equivalent- 0.00009 kg (3x10^8)^2
= 0.81x10^14 J
The mass converted to energy is called the mass defect. In this situation we have
mass defect = .003 % ( initial mass ) = .00003 * 3 kg = .00009 kg.
The energy equivalent of .00009 kg is
energy equivalent: E = m c^2 = .00009 kg ( 3 * 10^8 m/s) ^ 2 = .81 * 10^14 Joules.
This energy is released in the explosion.
Note that the energy in a kg of typical high explosives is around 3 * 10^7 J. In a metric ton, or 1000 kg, the energy would be approximately 3 * 10^10 J. So the energy in this explosion would be equivalent to on the order of 10,000 tons of high explosive.
This energy might also be released slowly in a nuclear power facility. One kilowatt-hour is approximately 3.6 * 10^6 Joules, so the energy released by this mass conversion would be on the order of 10^8 kilowatt-hours. At 7 cents per kwh, this energy would be worth on the order of ten million dollars.
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&#Good responses. Let me know if you have questions. &#
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