Bohr Magnetron Questions in English

(A) The magnetic moment $(\mu)$ of a revolving electron is given by the relation $\mu = \frac{e}{2m} L$,where $L$ is the angular momentum.
According to Bohr's quantization condition,the angular momentum $L$ of an electron in the $n$-th orbit is $L = \frac{nh}{2\pi}$.
Substituting this value into the expression for magnetic moment,we get $\mu = \frac{e}{2m} \left( \frac{nh}{2\pi} \right)$.
Rearranging the terms,we have $\mu = n \left( \frac{eh}{4\pi m} \right)$.
Since $e$,$h$,and $m$ are constants,the term $\frac{eh}{4\pi m}$ is also a constant (related to the Bohr magneton $\mu_B = \frac{eh}{4\pi m}$).
Therefore,$\mu \propto n$.

(B) The magnetic moment $\vec{M}$ of a current loop is given by $\vec{M} = I\vec{A}$,where $I$ is the current and $\vec{A}$ is the area vector.
For a proton of charge $e$ moving in a circular path of radius $r$ with speed $v$,the current $I$ is $I = \frac{e}{T} = \frac{e}{2\pi r / v} = \frac{ev}{2\pi r}$.
The area of the loop is $A = \pi r^2$.
Thus,the magnitude of the magnetic moment is $M = IA = \left(\frac{ev}{2\pi r}\right)(\pi r^2) = \frac{evr}{2}$.
The angular momentum of the proton is $L = m_pvr$.
Substituting $vr = L/m_p$ into the expression for $M$,we get $M = \frac{eL}{2m_p}$.
Since the proton is positively charged,the direction of the magnetic moment is the same as the direction of the angular momentum vector. Therefore,$\vec{M} = \frac{e\vec{L}}{2m_p}$.

(B) The magnetic moment $M$ of an electron revolving in a circular orbit is given by $M = I A$,where $I$ is the current and $A$ is the area of the orbit.
Since $I = \frac{e}{T}$ and $T = \frac{2\pi r}{v}$,we have $I = \frac{ev}{2\pi r}$.
The area $A = \pi r^2$.
Thus,$M = \left(\frac{ev}{2\pi r}\right) (\pi r^2) = \frac{1}{2} evr$.
We know that the angular momentum $J = mvr$.
Substituting $vr = \frac{J}{m}$ into the expression for $M$,we get:
$M = \frac{1}{2} e \left(\frac{J}{m}\right) = \frac{eJ}{2m}$.

(C) The magnetic moment $M$ of a current loop is given by $M = I A$,where $I$ is the current and $A$ is the area of the loop.
For an electron of charge $e$ moving in an orbit of radius $r$ with velocity $v$,the current $I = \frac{e}{T} = \frac{ev}{2\pi r}$.
The area of the orbit is $A = \pi r^2$.
Thus,$M = \left( \frac{ev}{2\pi r} \right) (\pi r^2) = \frac{evr}{2}$.
Multiplying and dividing by mass $m$,we get $M = \frac{e}{2m} (mvr)$.
According to Bohr's quantization condition,the angular momentum $L = mvr = \frac{nh}{2\pi}$.
Substituting this into the expression for $M$,we get $M = \left( \frac{e}{2m} \right) \frac{nh}{2\pi}$.

(N/A) Consider an electron of charge $(-e)$ performing uniform circular motion around a stationary heavy nucleus of charge $(+Ze)$ with orbital radius $r$ and speed $v$.
The circulating electron constitutes a current $I$ given by:
$I = \frac{e}{T} \quad \dots (1)$
where $T$ is the time period of revolution.
Since $v = \frac{2 \pi r}{T}$,we have $T = \frac{2 \pi r}{v} \quad \dots (2)$
Substituting $(2)$ into $(1)$:
$I = \frac{e v}{2 \pi r} \quad \dots (3)$
The magnetic moment $\mu_l$ associated with this current loop of area $A = \pi r^2$ is:
$\mu_l = I A = \left( \frac{e v}{2 \pi r} \right) (\pi r^2) = \frac{e v r}{2} \quad \dots (4)$
Multiplying and dividing by the electron mass $m_e$:
$\mu_l = \frac{e}{2 m_e} (m_e v r)$
Since the orbital angular momentum $L = m_e v r$,we get:
$\mu_l = \frac{e}{2 m_e} L$
The ratio $\frac{\mu_l}{L} = \frac{e}{2 m_e}$ is called the gyromagnetic ratio,which is a constant for an electron.

(N/A) Bohr magneton is defined as the magnetic moment associated with an electron due to its orbital motion in the first orbit of a hydrogen atom.
Bohr's first hypothesis states that the angular momentum of an electron in an orbit is quantized and given by:
$L = n \left( \frac{h}{2 \pi} \right)$
where $n = 1, 2, 3, \ldots$ and $h$ is Planck's constant $(h = 6.626 \times 10^{-34} \text{ J s})$.
The magnetic moment $\mu_l$ associated with an orbital electron is given by:
$\mu_l = \frac{e}{2 m_e} L$
Substituting the expression for angular momentum for the first orbit $(n = 1)$:
$\mu_l = \frac{e}{2 m_e} \left( \frac{h}{2 \pi} \right)$
This minimum value of the magnetic moment is called the Bohr magneton $(\mu_B)$:
$\mu_B = \frac{eh}{4 \pi m_e}$
Substituting the values $e = 1.6 \times 10^{-19} \text{ C}$,$h = 6.63 \times 10^{-34} \text{ J s}$,and $m_e = 9.11 \times 10^{-31} \text{ kg}$:
$\mu_B = \frac{(1.6 \times 10^{-19}) \times (6.63 \times 10^{-34})}{4 \times 3.14 \times 9.11 \times 10^{-31}}$
$\mu_B \approx 9.27 \times 10^{-24} \text{ A m}^2$

(N/A) Any charge in uniform circular motion has an associated magnetic moment given by the expression:
$\mu_{l} = \frac{e}{2m_{e}}(l)$
This dipole moment is known as the orbital magnetic moment. Its magnitude is related to the Bohr magneton,which has a value of $9.27 \times 10^{-24} \text{ A m}^2$.
In addition to the orbital magnetic moment,the electron possesses an intrinsic magnetic moment,which has the same numerical value of $9.27 \times 10^{-24} \text{ A m}^2$. This is called the spin magnetic moment.
It is important to note that this does not imply the electron is physically spinning. The electron is an elementary particle and does not have an axis to rotate around like a top or the Earth.

(N/A) An electron of charge $e$ revolving in a circular orbit of radius $r$ with a speed $v$ constitutes a current $I = \frac{e}{T}$,where $T$ is the time period of revolution.
Since $T = \frac{2\pi r}{v}$,the current is $I = \frac{ev}{2\pi r}$.
The orbital magnetic moment $\mu_l$ is given by the product of current $I$ and the area of the orbit $A = \pi r^2$.
$\mu_l = I \times A = \left( \frac{ev}{2\pi r} \right) \times (\pi r^2) = \frac{evr}{2}$.
In terms of angular momentum $L = mvr$,we can write $\mu_l = \frac{e}{2m} L$.

(A) The gyromagnetic ratio is defined as the ratio of the magnetic dipole moment $(\mu)$ of a particle to its angular momentum $(L)$.
Mathematically, it is expressed as: $\gamma = \frac{\mu}{L}$.
For an electron, the magnetic moment is $\mu = \frac{e}{2m} L$, where $e$ is the charge of the electron and $m$ is its mass.
Therefore, the gyromagnetic ratio for an electron is $\gamma = \frac{e}{2m}$.
The magnitude of the gyromagnetic ratio for an electron is approximately $8.8 \times 10^{10} \text{ C/kg}$.

The gyromagnetic ratio is defined as the ratio of the magnetic dipole moment $(M)$ of an electron to its angular momentum $(L)$.
Mathematically,it is expressed as: $\gamma = \frac{M}{L}$.
For an electron revolving in a circular orbit,the magnetic moment is $M = \frac{evr}{2}$ and the angular momentum is $L = mvr$.
Therefore,$\gamma = \frac{evr/2}{mvr} = \frac{e}{2m}$.
Substituting the values of the charge of an electron $(e \approx 1.6 \times 10^{-19} \ C)$ and the mass of an electron $(m \approx 9.1 \times 10^{-31} \ kg)$,the magnitude is:
$\gamma = \frac{1.6 \times 10^{-19}}{2 \times 9.1 \times 10^{-31}} \approx 8.8 \times 10^{10} \ C/kg$.

(N/A) $1$. Orbital Magnetic Moment: An electron revolving in an orbit around the nucleus behaves like a tiny current loop. The magnetic moment associated with this orbital motion is called the orbital magnetic moment. It is given by the formula $\mu_L = -\frac{e}{2m_e} L$,where $e$ is the charge of the electron,$m_e$ is the mass of the electron,and $L$ is the orbital angular momentum.
$2$. Intrinsic Magnetic Moment (Spin Magnetic Moment): Electrons possess an inherent property called spin,which is independent of their orbital motion. This spin gives rise to an intrinsic magnetic moment,often called the spin magnetic moment. It is associated with the intrinsic angular momentum (spin) of the electron and is approximately equal to one Bohr magneton,$\mu_B = \frac{eh}{4\pi m_e}$.

(B) The magnetic moment $\vec{\mu}$ of a current loop is given by $\vec{\mu} = I \vec{A}$.
For an electron of charge $-e$ revolving in an orbit of radius $r$ with speed $v$,the equivalent current is $I = \frac{-e}{T} = \frac{-ev}{2\pi r}$.
The area of the orbit is $A = \pi r^2$.
Thus,$\vec{\mu} = I \vec{A} = \left( \frac{-ev}{2\pi r} \right) (\pi r^2) = \frac{-evr}{2}$.
The orbital angular momentum is $\vec{L} = mvr$.
Substituting $vr = \frac{L}{m}$,we get $\vec{\mu} = -\frac{e}{2m} \vec{L}$.

(D) The Bohr magneton $(\mu_B)$ is defined as the magnetic moment of an electron orbiting in the first Bohr orbit of a hydrogen atom.
According to Bohr's quantization condition, the angular momentum $L$ is given by $L = m_e v r = \frac{n h}{2 \pi}$.
For the first orbit $(n = 1)$, $L = \frac{h}{2 \pi}$.
The magnetic moment $\mu$ of a current loop is given by $\mu = I A$, where $I = \frac{e}{T}$ and $T = \frac{2 \pi r}{v}$.
Thus, $\mu = \frac{e v}{2 \pi r} \times (\pi r^2) = \frac{e v r}{2}$.
Substituting $L = m_e v r$, we get $\mu = \frac{e L}{2 m_e}$.
Substituting $L = \frac{h}{2 \pi}$, we get $\mu_B = \frac{e}{2 m_e} \times \frac{h}{2 \pi} = \frac{e h}{4 \pi m_e}$.

(B) The magnetic moment $M$ is given by $M = I A$,where $I$ is the current and $A$ is the area of the orbit.
$I = \frac{e}{T} = \frac{ev}{2 \pi r}$ and $A = \pi r^2$.
Thus,$M = \left( \frac{ev}{2 \pi r} \right) (\pi r^2) = \frac{evr}{2}$.
Given the flux condition: $B(\pi r^2) = n(h/e)$. For the lowest energy state,$n = 1$,so $B \pi r^2 = h/e$,which implies $r^2 = \frac{h}{B \pi e}$.
Also,for an electron in a magnetic field,the centripetal force is provided by the Lorentz force: $\frac{mv^2}{r} = evB$,which gives $\frac{v}{r} = \frac{eB}{m}$.
Substituting $v = \frac{eBr}{m}$ into the expression for $M$:
$M = \frac{e}{2} \left( \frac{eBr}{m} \right) r = \frac{e^2 B r^2}{2m}$.
Substituting $r^2 = \frac{h}{B \pi e}$ into the equation for $M$:
$M = \frac{e^2 B}{2m} \left( \frac{h}{B \pi e} \right) = \frac{eh}{2 \pi m}$.

(B) The magnetic dipole moment $\overrightarrow{m}_{\text{orb}}$ of an electron revolving in a circular orbit is given by $\overrightarrow{m}_{\text{orb}} = -\frac{e}{2m} \overrightarrow{L}$,where $\overrightarrow{L}$ is the orbital angular momentum of the electron.
Taking the magnitude of both sides,we have $m_{\text{orb}} = \frac{e}{2m} L$.
Therefore,the ratio of angular momentum $L$ to the magnetic dipole moment $m_{\text{orb}}$ is given by $\frac{L}{m_{\text{orb}}} = \frac{2m}{e}$.

(A) The orbital magnetic moment $\vec{m}$ of an electron revolving in a circular orbit is given by $\vec{m} = I \vec{A}$,where $I$ is the current and $\vec{A}$ is the area vector. Since the electron is negatively charged,the direction of the equivalent current $I$ is opposite to the direction of the electron's motion. By the right-hand rule,the direction of $\vec{m}$ is perpendicular to the plane of the orbit.
The angular momentum $\vec{L}$ of the electron is given by $\vec{L} = \vec{r} \times \vec{p}$,where $\vec{p} = m_e \vec{v}$ is the linear momentum. According to the right-hand rule for the cross product,the direction of $\vec{L}$ is also perpendicular to the plane of the orbit.
Because the electron has a negative charge,the direction of the magnetic moment $\vec{m}$ is opposite to the direction of the angular momentum $\vec{L}$. Therefore,$\vec{m}$ and $\vec{L}$ are in opposite directions perpendicular to the plane of the orbit.

(C) The magnetic moment $\mu$ of a current loop is given by $\mu = iA$,where $i$ is the current and $A$ is the area of the loop.
For an electron revolving in a circular orbit of radius $r$ with period $T$,the equivalent current is $i = \frac{e}{T}$.
The area of the orbit is $A = \pi r^2$.
Thus,$\mu = \left(\frac{e}{T}\right) \pi r^2$.
The period $T$ is related to the orbital velocity $v$ by $T = \frac{2 \pi r}{v}$,so $\frac{1}{T} = \frac{v}{2 \pi r}$.
Substituting this into the expression for $\mu$,we get $\mu = e \left(\frac{v}{2 \pi r}\right) \pi r^2 = \frac{evr}{2}$.
The angular momentum of the electron is $L = mvr$,which implies $vr = \frac{L}{m}$.
Substituting $vr$ into the expression for $\mu$,we obtain $\mu = \frac{e}{2} \left(\frac{L}{m}\right) = \frac{e}{2m} L$.

(C) The orbital angular momentum $\vec{L}$ of an electron is given by $\vec{L} = \vec{r} \times \vec{p}$. For an electron revolving in a circular orbit,$\vec{L}$ is directed perpendicular to the plane of the orbit (upward,following the right-hand rule).
The orbital magnetic moment $\vec{M}$ is related to the orbital angular momentum $\vec{L}$ by the relation $\vec{M} = -\frac{e}{2m} \vec{L}$.
The negative sign indicates that the orbital magnetic moment $\vec{M}$ is in the direction opposite to the orbital angular momentum $\vec{L}$.
Since $\vec{M}$ and $\vec{L}$ are in opposite directions,the angle between them is $180^{\circ}$.

(A) The gyromagnetic ratio is defined as the ratio of the magnetic moment $(M_0)$ to the angular momentum $(L_0)$ of a particle.
Mathematically,it is expressed as:
$\text{Gyromagnetic ratio} = \frac{M_0}{L_0}$
For an electron revolving in an orbit,the magnetic moment $M_0 = \frac{e}{2m} L_0$,where $e$ is the charge and $m$ is the mass of the electron.
Thus,the ratio $\frac{M_0}{L_0} = \frac{e}{2m}$,which is a constant.

(D) The gyromagnetic ratio is defined as the ratio of the magnetic moment $(\mu_L)$ to the angular momentum $(L)$ of an orbital electron.
$\text{Gyromagnetic ratio} = \frac{\mu_L}{L} = \frac{e}{2m}$.
Given that the charge to mass ratio of an electron is $\frac{e}{m} = A \ C/kg$.
Substituting this into the expression,we get:
$\text{Gyromagnetic ratio} = \frac{1}{2} \times \left(\frac{e}{m}\right) = \frac{A}{2} \ C/kg$.

(C) Consider an electron of charge $e$ and mass $m$ revolving in a circular orbit of radius $r$ with a speed $v$ and time period $T$.
The current $I$ associated with this revolving electron is $I = \frac{e}{T} = \frac{e}{2\pi r / v} = \frac{ev}{2\pi r}$.
The magnetic moment $M$ is given by $M = I \times A = I \times (\pi r^2)$.
Substituting the value of $I$,we get $M = \left(\frac{ev}{2\pi r}\right) \times (\pi r^2) = \frac{evr}{2}$.
The angular momentum $L$ of the electron is $L = mvr$.
Thus,$vr = \frac{L}{m}$.
Substituting this into the expression for $M$,we get $M = \frac{e}{2} \times \left(\frac{L}{m}\right) = \frac{eL}{2m}$.

(B) The angular momentum of an electron in a circular orbit is given by $L = mvr$.
Since $v = r\omega$,we have $L = m(r\omega)r = m\omega r^2$.
Therefore,$\omega r^2 = \frac{L}{m} \quad \dots(i)$
The magnetic dipole moment $M$ associated with the revolving electron is given by $M = iA$.
Here,the current $i = \frac{e}{T} = \frac{e}{2\pi/\omega} = \frac{e\omega}{2\pi}$ and the area $A = \pi r^2$.
Substituting these values,$M = \left(\frac{e\omega}{2\pi}\right)(\pi r^2) = \frac{e}{2} \omega r^2$.
Substituting the value of $\omega r^2$ from equation $(i)$ into the expression for $M$:
$M = \frac{e}{2} \left(\frac{L}{m}\right)$.
Rearranging the terms to solve for $L$:
$L = \frac{2mM}{e}$.

(A) The gyromagnetic ratio is defined as the ratio of the magnetic dipole moment $(M)$ to the angular momentum $(L)$ of an electron orbiting in an atom.
For an electron in an orbit,the magnetic moment $M = I A = (\frac{e}{T}) (\pi r^2) = \frac{e}{2\pi r/v} (\pi r^2) = \frac{evr}{2}$.
The angular momentum $L = mvr$.
Thus,the gyromagnetic ratio $\gamma = \frac{M}{L} = \frac{evr/2}{mvr} = \frac{e}{2m}$.
The Bohr magneton $(\mu_B)$ is the fundamental unit of magnetic moment,given by $\mu_B = \frac{eh}{4\pi m}$.
Therefore,the gyromagnetic ratio and Bohr magneton are $\frac{e}{2m}$ and $\frac{eh}{4\pi m}$ respectively.

(A) The gyromagnetic ratio is defined as the ratio of the magnetic dipole moment $\mu$ to the angular momentum $L$ of an electron.
For an electron orbiting a nucleus,$\mu = \frac{e}{2m} L$.
Therefore,the gyromagnetic ratio $\gamma = \frac{\mu}{L} = \frac{e}{2m}$.
The specific charge of an electron is defined as $\frac{e}{m}$.
Comparing the two expressions,we get $\gamma = \frac{1}{2} \times (\frac{e}{m})$.
Thus,the gyromagnetic ratio is $1/2$ times the specific charge of an electron.

(B) The angular momentum of an electron in an orbit is given by $L = mvr$.
The magnetic dipole moment $M$ associated with the orbital motion of the electron is given by $M = IA$,where $I$ is the current and $A$ is the area of the orbit.
Since $I = \frac{e}{T} = \frac{ev}{2 \pi r}$ and $A = \pi r^2$,we have $M = \left( \frac{ev}{2 \pi r} \right) \times (\pi r^2) = \frac{1}{2} evr$.
Taking the ratio of the magnetic dipole moment to the angular momentum:
$\frac{M}{L} = \frac{\frac{1}{2} evr}{mvr} = \frac{e}{2m}$.
Thus,the ratio is $\frac{e}{2m}$.

(D) The gyromagnetic ratio is defined as the ratio of the magnetic dipole moment to the angular momentum of the electron,which is given by the formula: $\gamma = \frac{e}{2m_e}$.
Given,gyromagnetic ratio $\gamma = 8.8 \times 10^{10} \ C \ kg^{-1}$.
Charge of the electron $e = 1.6 \times 10^{-19} \ C$.
Rearranging the formula to solve for the mass of the electron $m_e$:
$m_e = \frac{e}{2\gamma}$.
Substituting the given values:
$m_e = \frac{1.6 \times 10^{-19}}{2 \times 8.8 \times 10^{10}}$.
$m_e = \frac{1.6 \times 10^{-19}}{17.6 \times 10^{10}}$.
$m_e = \frac{16}{176} \times 10^{-29} \ kg$.
$m_e = \frac{1}{11} \times 10^{-29} \ kg$.
Thus,the mass of the electron is $\frac{1}{11} \times 10^{-29} \ kg$.

(B) When an electron orbits around the hydrogen nucleus,it constitutes a current loop.
The current $i$ is given by $i = \frac{e}{T}$,where $T$ is the time period of revolution.
The magnetic moment is $\mu = i A = \frac{e}{T} (\pi r^2)$.
The time period $T = \frac{2 \pi r}{v}$. Substituting this into the expression for $\mu$:
$\mu = \frac{e v}{2 \pi r} (\pi r^2) = \frac{e v r}{2}$.
The orbital angular momentum is $L = m_e v r$,which implies $v r = \frac{L}{m_e}$.
Substituting $v r$ into the expression for $\mu$:
$\mu = \frac{e}{2} \left(\frac{L}{m_e}\right) = \frac{e}{2 m_e} L$.
Since the electron is negatively charged,the magnetic moment vector $\vec{\mu}$ and angular momentum vector $\vec{L}$ are in opposite directions:
$\vec{\mu} = -\left(\frac{e}{2 m_e}\right) \vec{L}$.

(C) The magnetic moment $M$ of a particle of charge $q$ moving in a circular orbit of radius $r$ with speed $v$ is given by $M = I A = (\frac{q}{T}) (\pi r^2) = (\frac{q \omega}{2 \pi}) (\pi r^2) = \frac{1}{2} q \omega r^2$.
The angular momentum $L$ of the particle is given by $L = mvr = m(\omega r)r = m \omega r^2$.
The ratio of the magnitude of magnetic moment to angular momentum is $\frac{M}{L} = \frac{\frac{1}{2} q \omega r^2}{m \omega r^2} = \frac{q}{2m}$.
This ratio is known as the gyromagnetic ratio.

(C) The magnetic moment $M$ of a particle of charge $q$ moving in a circular orbit is given by $M = IA$,where $I$ is the current and $A$ is the area of the orbit.
The current $I = \frac{q}{T} = \frac{q\omega}{2\pi}$.
The area $A = \pi r^2$.
Thus,$M = \left(\frac{q\omega}{2\pi}\right)(\pi r^2) = \frac{q\omega r^2}{2}$.
The angular momentum $L$ of the particle is $L = mvr = m(\omega r)r = m\omega r^2$.
The ratio of the magnetic moment to the angular momentum is $\frac{M}{L} = \frac{q\omega r^2 / 2}{m\omega r^2} = \frac{q}{2m}$.
This ratio is known as the gyromagnetic ratio and depends only on the charge '$q$' and mass '$m$' of the particle.

(C) The current $I$ is given by $I = \frac{e}{T} = \frac{e \omega}{2 \pi}$. This is independent of $r$.
The magnetic moment $\mu = I A = \left( \frac{e \omega}{2 \pi} \right) (\pi r^2) = \frac{e \omega r^2}{2}$.
The angular momentum $L = m v r = m (\omega r) r = m \omega r^2$.
Comparing $\mu$ and $L$,we get $\mu = \frac{e}{2m} L$.
Thus,the magnetic moment is proportional to the angular momentum,and the ratio $\mu / L = e / (2m)$ is known as the gyromagnetic ratio. Statement $C$ is correct as it relates $\mu$ and $L$ via the factor $e / (2m)$.