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Chapter 41, Problem 4P

(a)

To determine

Show that the electron in classical hydrogen atom spirals into nucleus at a rate drdt=e412π2ε02me2c3(1r2).

(a)

Expert Solution
Check Mark

Answer to Problem 4P

The electron in classical hydrogen atom spirals into nucleus at a rate drdt=e412π2ε02me2c3(1r2).

Explanation of Solution

Write the equation for acceleration according to Newton’s second law.

  a=Fm                                                                                           (I)

Here, F is the force, a is the acceleration and m is the mass.

Write the equation for the force experienced by the particle moving in a uniform circular motion.

  F=kee2r2                                                                                         (II)

Here, ke is the Coulomb constant, e is the electric charge and r is the radius.

Write the acceleration equation using the above equation and substitute 1/4πε0 for ke.

  a=kee2mer2=e24πε0mer2                                                                                 (III)

Here, me is the mass of the electron and ε0 is the permittivity of free space.

Write the equation for centripetal acceleration.

  a=v2r                                                                                (IV)

Here, v is the velocity of the rotating particle.

Compare equations (III) and (IV) and rearrange to find mev2.

  e24πε0mer2=v2rmev2=e24πε0r                                                                     (V)

Write the equation for the total energy.

  E=K+U                                                                               (VI)

Here, E is the total energy, K is the kinetic energy and U is the potential energy.

Write the equation for kinetic energy.

  K=12mev2                                                                               (VII)

Write the equation for potential energy.

  U=e28πε0r                                                                            (VIII)

Substitute the equation (VII) and (VIII) in equation (VI) and use equation (V) to find E.

  E=12mev2e24πε0r=e28πε0re24πε0r=e28πε0r                                                                      (IX)

Conclusion:

Write the time given time derivative equation for energy.

  dEdt=16πε0e2a2c3

Substitute the equation (III) and (IX) in the above equation.

  d(e28πε0r)dt=16πε0e2c3(e24πε0mer2)2e28πε0r2drdt=16πε0e2c3(e24πε0mer2)2drdt=e412π2ε02me2c3(1r2)

Thus, the electron in classical hydrogen atom spirals into nucleus at a rate drdt=e412π2ε02me2c3(1r2).

(b)

To determine

The time interval over which the electron reaches r=0.

(b)

Expert Solution
Check Mark

Answer to Problem 4P

The time interval over which the electron reaches r=0 is 0.846ns.

Explanation of Solution

Write the equation for the rate at which the electron in classical hydrogen atom spirals into nucleus.

  drdt=e412π2ε02me2c3(1r2)dt=12π2ε02me2c3r2e4dr                                                                  (X)

Take the integral of the time.

  T=0Tdt                                                                                         (XI)

Here, T is the time interval over which the electron reaches r=0.

Substitute equation (X) in (XI) to find T.

  T=2.00×1010m012π2ε02me2c3r2e4dr=02.00×1010m12π2ε02me2c3r2e4dr=12π2ε02me2c3r33e4|02.00×1010m                                                        (XII)

Conclusion:

Substitute 8.85×1012C for ε0, 9.11×1031kg for me, 3.00×108m/s for c and 1.60×1019C for e in the above equation.

  T=12π2(8.85×1012C)2(9.11×1031kg)(3.00×108m/s)3(2.00×1010m)33(1.60×1019C)4=(0.846×109s)(109ns1s)=0.846ns

Thus, the time interval over which the electron reaches r=0 is 0.846ns.

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