
What is the correct value of Bohr magneton?
\[
A.8.99 \times {10^{ - 24}}A{m^2} \\
B.9.27 \times {10^{ - 24}}A{m^2} \\
C.5.66 \times {10^{ - 24}}A{m^2} \\
D.9.27 \times {10^{ - 28}}A{m^2} \\
\]
Answer
578.7k+ views
Hint: Bohr magneton is used in atomic physics. A physical constant used to express the magnetic moment of an electron caused due to spin or orbital angular momentum. The magnetic moment of the charged particle can be calculated when moving charge forms current so that orbital motion of the electron generates the magnetic moment. The other way is by spin of the electron giving a magnetic moment.
Complete step-by-step answer:
Bohr magneton: The smallest value of orbital magnetic moment of the electron for n=1 is given by
$M = \dfrac{{e\hbar }}{{2m}} = \dfrac{{eh}}{{4\pi m}}$
This is the unit of magnetic moment which is known as a Bohr magneton. It is denoted by μB.
Where,
e is the electron with charge \[1.6{\text{ }} \times {\text{ }}{10^{ - 19}}C\]
h is the Planck’s constant \[6.6{\text{ }} \times {\text{ }}{10^{ - 34}}Js\]
m is the mass of the electron in motion \[9.1{\text{ }} \times {\text{ }}{10^{ - 31}}kg\]
Substituting the known data in the above equation, we will get
$
{\mu _B} = \dfrac{{eh}}{{4\pi m}} = \dfrac{{1.6 \times {{10}^{ - 19}} \times 6.6 \times {{10}^{ - 34}}}}{{4\pi \times 9 \times {{10}^{ - 31}}}} \\
{\mu _B} = 9.27 \times {10^{ - 24}}A{m^2} \\
$
This is the value of Bohr Magneton and the correct option is B.
Note: The unit of Bohr magneton is also the same as that of M. As M=IA, the unit of Bohr magneton is also Am-2. Bohr magneton is useful as a unit of measurement to measure atomic magnetic moments.
Complete step-by-step answer:
Bohr magneton: The smallest value of orbital magnetic moment of the electron for n=1 is given by
$M = \dfrac{{e\hbar }}{{2m}} = \dfrac{{eh}}{{4\pi m}}$
This is the unit of magnetic moment which is known as a Bohr magneton. It is denoted by μB.
Where,
e is the electron with charge \[1.6{\text{ }} \times {\text{ }}{10^{ - 19}}C\]
h is the Planck’s constant \[6.6{\text{ }} \times {\text{ }}{10^{ - 34}}Js\]
m is the mass of the electron in motion \[9.1{\text{ }} \times {\text{ }}{10^{ - 31}}kg\]
Substituting the known data in the above equation, we will get
$
{\mu _B} = \dfrac{{eh}}{{4\pi m}} = \dfrac{{1.6 \times {{10}^{ - 19}} \times 6.6 \times {{10}^{ - 34}}}}{{4\pi \times 9 \times {{10}^{ - 31}}}} \\
{\mu _B} = 9.27 \times {10^{ - 24}}A{m^2} \\
$
This is the value of Bohr Magneton and the correct option is B.
Note: The unit of Bohr magneton is also the same as that of M. As M=IA, the unit of Bohr magneton is also Am-2. Bohr magneton is useful as a unit of measurement to measure atomic magnetic moments.
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