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Classically, an orbiting charged particle radiates electromagnetic energy, and for an electron in atomic dimensions, it would lead to collapse in considerably less than the wink of an eye.

(a) By equating the centripetal and Coulomb forces, show that for a classical charge -e of mass m held in a circular orbit by its attraction to a fixed charge +e, the following relationship holds

ω=er-3/24πε0m.

(b) Electromagnetism tells us that a charge whose acceleration is a radiates power P=e2a2/6ε0c3. Show that this can also be expressed in terms of the orbit radius as P=e696π2ε03m2c3r4. Then calculate the energy lost per orbit in terms of r by multiplying the power by the period T=2π/ωand using the formula from part (a) to eliminate .

(c) In such a classical orbit, the total mechanical energy is half the potential energy, or Eorbit=-e28πε0r. Calculate the change in energy per change in r : dEorbit/dr. From this and the energy lost per obit from part (b), determine the change in per orbit and evaluate it for a typical orbit radius of 10-10m. Would the electron's radius change much in a single orbit?

(d) Argue that dividing dEorbit/dr by P and multiplying by dr gives the time required to change r by dr . Then, sum these times for all radii from rinitial to a final radius of 0. Evaluate your result for rinitial=10-10m. (One limitation of this estimate is that the electron would eventually be moving relativistically).

Short Answer

Expert verified

(a) It is proved that ω=er-3/24πε0m.

(b) It is proved that the radiated power ise696π2ε03m2c3r4. Energy lost per orbit ise524πε05/2m3/2c3r5/2.

(c) The change in radius per orbit is 0.39×10-10mwhich is not small compared to the actual radius.

(d) The time taken to change the radius from its initial value to zero is 3.36×10-11s.

Step by step solution

01

Given data

An electron of charge -e is orbiting around a positive charge e in a circular orbit of radius r.

The total energy of the electron is

Eorbit=-e28πε0r.....(I)

Here ε0is the permittivity of free space.

The power of electromagnetic radiation radiated by an electron with acceleration a is

P=e2a2/6ε0c3.....(II)

Here c is the speed of light in vacuum.

02

Coulomb force, centripetal force and centripetal acceleration

The Coulomb force between two charge particles of charge q and distance r is

F=q24πε0r .....(III)

The centripetal force of a particle of mass m moving in circular orbit of radius r and angular velocity ωis

F=mω2r.....(IV)

The angular acceleration of a particle undergoing uniform circular motion of radius r and angular velocity ω is

a=ω2r.....(V)

03

Determining the angular velocity

Equate equations (III) for q=eand (IV) to get

e24πε0r=mω2rω2=e24πε0mr3ω=e4πε0mr3/2

04

Determining the power and energy lost per orbit

Substitute equation (V) in (II) and use form of angular velocity to get

P=e2r2ω46ε0c3=e2r26ε0c3×e416π2ε02m2r6=e696π2ε03m2c3r4

To get the energy lost per orbit multiply this by the time period T=2π/ω

ΔE=P×T=e696π2ε03m2c3r4×2πω=e648πε03m2c3r44πε0mr3/2e=e524πε05/2m3/2c3r5/2

05

Determining the change in radius per orbit

Differentiate equation (I) with respect to r and get

dEorbitdr=e28πε0r2dr=8πε0r2×dEorbite2

Substitute expression of change in energy per orbit from the previous section to get

r=8πε0r2×ΔEorbite2=8πε0r2e2×e524πε05/2m3/2c3r5/2=πe33ε03/2m3/2c3r

Substitute the values

e=1.6×10-19Cε0=8.85×10-12C2/N·m2m=9.1×10-31kgr=10-10mc=3×108m/sandget

r=3.14×1.6×10-19C33×8.85×10-12C2/N·m23/2×9.1×10-31kg3/2×3×108m/s3×10-10m=0.0039×10-8×1C3·11C3·1N3/2×1kg3/2·m3/2/s31N3/2·1m3·11kg3/2·11m3/s3·1m-1/2=0.39×10-10m

Thus the change in radius per orbit is 0.39×10-10m.

06

Determining the time required to change radius

dEorbit/dris the change in energy per unit change in radius and power is the change in energy per unit time. Dividing the two will give the inverse of change in radius of time. Thus multiplying with dr will give the time taken to change radius by dr .

The expression will be

dt=1PdEorbitdrdr=96π2ε03m2c3r4e6e28πε0r2dr=12πε02m2c3e4r2dr

Integrate this from rinitial=10-10mto 0 to get

Δt=12πε02m2c3e4rinitial0r2dr=12πε02m2c3rinitial33e4=4πε02m2c3rinitial3e4

Substitute the values to get

Δt=4π8.85×10-12C2/N·m229.1×10-31kg23×108m/s310-10m31.6×10-19C4=3.36×10-11·1C4·11N2×1N2s41kg2m2·11m4·1kg2·1m3/s3·1m3·11C4=3.36×10-11s

Thus the time taken is 3.36×10-11s.

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Most popular questions from this chapter

Classically, it was expected that an orbiting electron would emit radiation of the same frequency as its orbit frequency. We have often noted that classical behaviour is observed in the limit of large quantum numbers. Does it work in this case? (a) Show that the photon energy for the smallest possible energy jump at the “low-n-end” of the hydrogen energies is 3|E0|/n3, while that for the smallest jump at the “high-n-end” is 2|E0|/n3, where E0is hydrogen’s ground-state energy. (b) Use F=ma to show that the angular velocity of a classical point charge held in orbit about a fixed-point charge by the coulomb force is given by ω=e2/4πε0mr3. (c) Given that r=n2a0, is this angular frequency equal to the minimum jump photon frequency at either end of hydrogen’s allowed energies?

What is a quantum number, and how does it arise?

Knowing precisely all components of a nonzero Lwould violate the uncertainty principle, but knowingthat Lis precisely zerodoes not. Why not?

(Hint:For l=0 states, the momentum vector p is radial.)

A hydrogen atom in an n = 2 state absorbs a photon,

  1. What should be the photon wavelength to cause the electron to jump to an n = 4 state?
  2. What wavelength photons might be emitted by the atom following this absorption?

For a hydrogen atom in the ground state. determine (a) the most probable location at which to find the electron and (b) the most probable radius at which to find the electron, (c) Comment on the relationship between your answers in parts (a) and (b).

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