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Question: Bohr Magneton value in SI unit is: A.\(9.273 \times {10^{ - 24}}erg{T^{ - 1}}\) B.\(9.273 \times...

Bohr Magneton value in SI unit is:
A.9.273×10−24ergT−19.273 \times {10^{ - 24}}erg{T^{ - 1}}
B.9.273×10−24JT−19.273 \times {10^{ - 24}}J{T^{ - 1}}
C.9.273×10−17JT−19.273 \times {10^{ - 17}}J{T^{ - 1}}
D.9.273×10−10calT−19.273 \times {10^{ - 10}}cal{T^{ - 1}}

Explanation

Solution

The magnetic moment of the electron is due to its orbital momentum or spin orbital momentum. The Bohr Magneton is the physical constant named for the magnetic moment created by spin angular momentum or orbital momentum which is denoted by B.
The SI unit of Bohr Magneton is given by equation as shown below.
μB=eℏ2me{\mu _B} = \dfrac{{e\hbar }}{{2{m_e}}}
Where
e is the elementary charge.
ℏ\hbar is reduced planck constant.
me{m_e} is the reduced mass of the electron.

Complete step by step answer:
The electron moves in the circular path with the velocity v and having the radius r.
The time is given by the equation as shown below.
T=2πrvT = \dfrac{{2\pi r}}{v}
As the electron moves in the circular path, the charge is develop which is given by the equation as shown below
I=−ev2πrI = \dfrac{{ - ev}}{{2\pi r}}
The magnetic moment is given by the equation as shown below
m=IAm = IA
Where, A is area of the circular path.
μ=IA\mu = IA
⇒μ=−ev2πr×πr2\Rightarrow \mu = \dfrac{{ - ev}}{{2\pi r}} \times \pi {r^2}
⇒μ=−erv2\Rightarrow \mu = \dfrac{{ - erv}}{2}
Divide and multiply the above equation with mass of electron
μ=−e2memevr\mu = \dfrac{{ - e}}{{2{m_e}}}{m_e}vr
In this equation, mevr{m_e}vr is the angular momentum which is represented by L.
μ=−e2meL.......(i)\mu = \dfrac{{ - e}}{{2{m_e}}}L.......(i)
When an electron moves in a circular orbit, the angular momentum will be quantized and is given by
L=nh2πL = \dfrac{{nh}}{{2\pi }}
Substitute, the value of L in equation (i)
μ=−e2me×nh2π\mu = \dfrac{{ - e}}{{2{m_e}}} \times \dfrac{{nh}}{{2\pi }}
⇒μ=−neh4πme\Rightarrow \mu = - n\dfrac{{eh}}{{4\pi {m_e}}}
Here, eh4πme\dfrac{{eh}}{{4\pi {m_e}}} is the Bohr magneton
μB=eh4πme{\mu _B} = \dfrac{{eh}}{{4\pi {m_e}}}
⇒μB=eℏ2me\Rightarrow {\mu _B} = \dfrac{{e\hbar }}{{2{m_e}}}
⇒μB=9.27×10−24J/T\Rightarrow {\mu _B} = 9.27 \times {10^{ - 24}}J/T
Thus, the value of Bohr magneton is9.274×10−24J/T9.274 \times {10^{ - 24}}J/T or 9.274×10−249.274 \times {10^{ - 24}}joules per tesla.
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Therefore, the correct option is B.

Note: The CGS unit of Bohr magneton is given by the equation as shown below.
μB=eℏ2mec{\mu _B} = \dfrac{{e\hbar }}{{2{m_e}c}}
Where
e is the elementary charge.
ℏ\hbar is the reduced planck constant.
me{m_e} is the reduced mass of the electron.
c is the speed of light