The angular velocity of the electron in the lowest Bohr orbit of the hydrogen atom is approximately \(4 \times 10^{16} \mathrm{~s}^{-1}\). What is the largest magnetic field which may be regarded as small in this case, in the sense of \(\S 5.5 ?\) Determine the Larmor frequency in a field of \(2 \mathrm{~T}\).

We have to find the magnetic moment of the electron first. The magnetic moment of an electron is given by the formula: \(\boldsymbol{\mu} = -\dfrac{e}{2m_{e}}\boldsymbol{L}\) where \(e\) is the charge of the electron, \(m_{e}\) is the mass of the electron, and \(\boldsymbol{L}\) is the angular momentum of the electron. The angular momentum of the electron is related to its angular velocity by the equation: \(\boldsymbol{L} = m_{e}r^2\boldsymbol{\omega}\) where \(r\) is the radius of the lowest Bohr orbit and \(\boldsymbol{\omega}\) is the angular velocity. Given the angular velocity of the electron, we can calculate the magnetic moment as follows: \(\boldsymbol{\mu} = -\dfrac{e}{2m_{e}}(m_{e}r^2\boldsymbol{\omega})\)

Now we need to find the largest magnetic field under which the electron's magnetic moment will still precess slowly (in the sense of section 5.5). In order for the magnetic field to be considered small, the Larmor frequency should be much smaller than the electron's angular frequency, which corresponds to the condition \(\omega_{L} \ll \boldsymbol{\omega}\) where \(\omega_{L}\) is the Larmor frequency. The Larmor frequency is given by the formula: \(\omega_{L} = \dfrac{|\boldsymbol{\mu}|B}{\hbar}\) Substituting the magnetic moment expression, we get: \(\omega_{L} = \dfrac{-e}{2m_e}(m_{e}r^2\boldsymbol{\omega})B \div \hbar\) For \(\omega_{L} \ll \boldsymbol{\omega}\), we can write the condition: \(B \ll \dfrac{2\hbar\boldsymbol{\omega}}{-e(r^2)}\) Now we have all the quantities we need to find the largest magnetic field that can be considered small. Substituting the given angular velocity and the values of the constants \(e\) (elementary charge), \(\hbar\) (reduced Planck constant), and \(r\) (Bohr radius), we can calculate the largest magnetic field: \(B \ll \dfrac{2(1.054 \times 10^{-34}\,\mathrm{J\cdot s})(4 \times 10^{16}\,\mathrm{s}^{-1})}{1.602 \times 10^{-19}\,\mathrm{C}(5.29 \times 10^{-11}\,\mathrm{m})^2}\) \(B \ll 1.96 \times 10^{-4} \,\mathrm{T}\) So, the largest magnetic field that may be regarded as small is \(1.96 \times 10^{-4} \,\mathrm{T}\).

Now we need to find the Larmor frequency of the electron in a magnetic field of strength \(2\,\mathrm{T}\). We can use the Larmor frequency equation derived in step 2: \(\omega_{L} = \dfrac{-e}{2m_e}(m_{e}r^2\boldsymbol{\omega})B \div \hbar\) Substituting the given magnetic field strength and the values of the constants \(e\), \(\hbar\), \(\boldsymbol{\omega}\), and \(r\), we can calculate the Larmor frequency: \(\omega_{L} = \dfrac{-1.602 \times 10^{-19}\,\mathrm{C}}{2(9.109 \times 10^{-31}\,\mathrm{kg})}((9.109 \times 10^{-31}\,\mathrm{kg})(5.29 \times 10^{-11}\,\mathrm{m})^2(4 \times 10^{16}\,\mathrm{s}^{-1}))(2\,\mathrm{T}) \div (1.054 \times 10^{-34}\,\mathrm{J\cdot s})\) \(\omega_{L} = 2.8 \times 10^{11}\,\mathrm{s^{-1}}\) So, the Larmor frequency of the electron in a \(2\,\mathrm{T}\) magnetic field is \(2.8 \times 10^{11}\,\mathrm{s^{-1}}\).