Magnetic Field of a Hydrogen Atom
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The following quantities appear in the problem:
Zeit \(t\) / Masse \(m\) / elektrische Stromstärke \(I\) / elektrische Ladung \(q, Q\) / Magnetische Flussdichte \(B\) / Kraft \(F\) / Geschwindigkeit \(v\) / Strecke \(s\) / Radius \(r\) / Umfang \(u\) /
The following formulas must be used to solve the exercise:
\(F = \frac{1}{4\pi\epsilon_0}\cdot \frac{q_1q_2}{r^2} \quad \) \(I = \dfrac{q}{t} \quad \) \(s = vt \quad \) \(u = 2\pi r \quad \) \(B = \dfrac{\mu_0 I}{2r} \quad \) \(F = m\dfrac{v^2}{r} \quad \)
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Exercise:
According to the Bohr model of the hydrogen atom the electron orbits the nucleus on a circular trajectory with radius rBO em Bohr radius for ground state. This can be erpreted as a circular current loop. Calculate the corresponding magnetic field.
Solution:
The orbital frequency can be found using the centripetal condition with the electrostatic force acting as the centripetal force: sscFC fracpiepsilon_fracq_p q_er^ frace^piepsilon_ r^ m_e omega^ r pi^ m_e f^ r Solving for the frequency f leads to f sqrtfrace^pi^epsilon_ m_e r^ The current produced by the electron corresponds to I fracDelta QDelta t fraceT e f sqrtfrace^pi^epsilon_ m_e r^ The magnetic field at the centre of the atom is therefore given by B fracmu_ fracIr BF fracncmuotimessqrtfracnce^pi^timesncepstimes metimes rB^ B approx resultBP vspacecm Remark: The magnetic field at the centre of the hydrogen atom is very strong. This does not imply though that we can expect a very strong macroscopic field. A better measure for the strength of a dipole magnet is the magnetic dipole moment. For a circular loop with area A carrying a current I the dipole moment is mu I A For the hydrogen atom we find mu I A sqrtfrace^pi^epsilon_ m_e r^ r^ pi dmF sqrtfracnce^ times rBpi times nceps times ncme dmS A typical value for the dipole moment of a bar magnet is dmtypO.
According to the Bohr model of the hydrogen atom the electron orbits the nucleus on a circular trajectory with radius rBO em Bohr radius for ground state. This can be erpreted as a circular current loop. Calculate the corresponding magnetic field.
Solution:
The orbital frequency can be found using the centripetal condition with the electrostatic force acting as the centripetal force: sscFC fracpiepsilon_fracq_p q_er^ frace^piepsilon_ r^ m_e omega^ r pi^ m_e f^ r Solving for the frequency f leads to f sqrtfrace^pi^epsilon_ m_e r^ The current produced by the electron corresponds to I fracDelta QDelta t fraceT e f sqrtfrace^pi^epsilon_ m_e r^ The magnetic field at the centre of the atom is therefore given by B fracmu_ fracIr BF fracncmuotimessqrtfracnce^pi^timesncepstimes metimes rB^ B approx resultBP vspacecm Remark: The magnetic field at the centre of the hydrogen atom is very strong. This does not imply though that we can expect a very strong macroscopic field. A better measure for the strength of a dipole magnet is the magnetic dipole moment. For a circular loop with area A carrying a current I the dipole moment is mu I A For the hydrogen atom we find mu I A sqrtfrace^pi^epsilon_ m_e r^ r^ pi dmF sqrtfracnce^ times rBpi times nceps times ncme dmS A typical value for the dipole moment of a bar magnet is dmtypO.