
Derive an expression for the magnetic moment of an electron revolving around the nucleus in terms of its angular momentum. What is the direction of the magnetic moment of an electron with respect to its angular momentum?
Answer
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Hint:Magnetic moment is current multiplied by area of loop. Current is charge per unit time and time period is distance covered divided by time. So, substitute the value of time period in the current equation and substitute this value of current in the formula of magnetic moment.
Complete step by step answer:
The magnetic moment is a determination of its tendency to get arranged by the influence of a magnetic field. The direction of magnetic moment points from south to north in a bar magnet. The SI unit of magnetic dipole moment is $Wb\,m$ and the SI unit of magnetic moment is $A{{m}^{2}}$ or J$/$Tesla.
The electron will revolve around the nucleus in a circular path. So, consider an electron moving around the nucleus in a circular path having radius $r$ and with velocity $v$. Current is charge per unit time: $i=\dfrac{q}{t}$
The charge of an electron is
$e=1.6\times {{10}^{-19}}$C.
So, $I=\dfrac{e}{t}$ where $t$ is the time period.
Formula of time period is:
$t=\dfrac{2\pi r}{v}$
Now, substituting value of t, we get
$i=\dfrac{ev}{2\pi r}$
Area of current loop (i.e. electron orbit)
$A=\pi {{r}^{2}}$
Magnetic moment due to motion of electron is:
$M=I\times A=\dfrac{ev}{2\pi r}\times \pi {{r}^{2}}$
$M=\dfrac{evr}{2}$
The direction of magnetic moment is along the x-axis.Momentum is the product of mass and velocity of the object. Any object in motion having some mass possesses momentum. The only difference between linear momentum and angular momentum is that it deals with rotating objects.
Formula of angular momentum is:
$L=mvr$
where $m$ is mass, $v$ is velocity and $r$ is radius of path.
SI unit of angular momentum is $kg{{m}^{2}}{{s}^{-1}}$.
Angular momentum of an electron is $L={{m}_{e}}vr$ where ${{m}_{e}}$ is mass of electron. Now to establish a relation between magnetic moment and angular momentum divide magnetic moment by angular momentum.
$\dfrac{M}{L}=\dfrac{{evr}/{2}\;}{{{m}_{e}}vr}=\dfrac{e}{2{{m}_{e}}}$
Thus, magnetic moment is $M=\dfrac{e}{2{{m}_{e}}}L$
In vector form $M=\dfrac{-e}{2{{m}_{e}}}\times L$.
Negative sign is used as the direction of magnetic moment is always opposite to that of the direction of angular momentum. Also, it is positive when the movement is anticlockwise and negative when the movement is clockwise.
Note: The charge of an electron is $e=-1.6\times {{10}^{-19}}$C and mass of electron is ${{m}_{e}}=9.1\times {{10}^{-31}}$kg. The stationary orbits in which electrons are revolving around the nucleus in the atom are not circular but elliptical in shape. It is due to the influence of the centrally located nucleus.
Complete step by step answer:
The magnetic moment is a determination of its tendency to get arranged by the influence of a magnetic field. The direction of magnetic moment points from south to north in a bar magnet. The SI unit of magnetic dipole moment is $Wb\,m$ and the SI unit of magnetic moment is $A{{m}^{2}}$ or J$/$Tesla.
The electron will revolve around the nucleus in a circular path. So, consider an electron moving around the nucleus in a circular path having radius $r$ and with velocity $v$. Current is charge per unit time: $i=\dfrac{q}{t}$
The charge of an electron is
$e=1.6\times {{10}^{-19}}$C.
So, $I=\dfrac{e}{t}$ where $t$ is the time period.
Formula of time period is:
$t=\dfrac{2\pi r}{v}$
Now, substituting value of t, we get
$i=\dfrac{ev}{2\pi r}$
Area of current loop (i.e. electron orbit)
$A=\pi {{r}^{2}}$
Magnetic moment due to motion of electron is:
$M=I\times A=\dfrac{ev}{2\pi r}\times \pi {{r}^{2}}$
$M=\dfrac{evr}{2}$
The direction of magnetic moment is along the x-axis.Momentum is the product of mass and velocity of the object. Any object in motion having some mass possesses momentum. The only difference between linear momentum and angular momentum is that it deals with rotating objects.
Formula of angular momentum is:
$L=mvr$
where $m$ is mass, $v$ is velocity and $r$ is radius of path.
SI unit of angular momentum is $kg{{m}^{2}}{{s}^{-1}}$.
Angular momentum of an electron is $L={{m}_{e}}vr$ where ${{m}_{e}}$ is mass of electron. Now to establish a relation between magnetic moment and angular momentum divide magnetic moment by angular momentum.
$\dfrac{M}{L}=\dfrac{{evr}/{2}\;}{{{m}_{e}}vr}=\dfrac{e}{2{{m}_{e}}}$
Thus, magnetic moment is $M=\dfrac{e}{2{{m}_{e}}}L$
In vector form $M=\dfrac{-e}{2{{m}_{e}}}\times L$.
Negative sign is used as the direction of magnetic moment is always opposite to that of the direction of angular momentum. Also, it is positive when the movement is anticlockwise and negative when the movement is clockwise.
Note: The charge of an electron is $e=-1.6\times {{10}^{-19}}$C and mass of electron is ${{m}_{e}}=9.1\times {{10}^{-31}}$kg. The stationary orbits in which electrons are revolving around the nucleus in the atom are not circular but elliptical in shape. It is due to the influence of the centrally located nucleus.
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