
What is the ratio of Bohr magneton to the nuclear magneton?
\[\begin{align}
& \text{A}\text{.}\dfrac{{{m}_{p}}}{{{m}_{e}}} \\
& \text{B}\text{.}\dfrac{m_{p}^{2}}{m_{e}^{2}} \\
& \text{C}\text{.1} \\
& \text{D}\text{.}\dfrac{{{m}_{e}}}{{{m}_{p}}} \\
\end{align}\]
Answer
509.5k+ views
Hint: Bohr Magneton (${{\mu }_{B}}$) is the magnetic moment of an electron caused by either its spin or angular momentum. Nuclear magneton (${{\mu }_{N}}$) is the magnetic dipole moment for heavier particles like protons, atomic nuclei etc. Their equations are different only in terms of mass. Bohr magneton uses electron mass while nuclear magneton uses proton mass.
Formula used:
\[{{\mu }_{B}}=\dfrac{e\hbar }{2{{m}_{e}}}\]
\[{{\mu }_{N}}=\dfrac{e\hbar }{2{{m}_{p}}}\]
Complete step by step solution:
Bohr Magneton is the magnetic moment of an electron caused by its spin or angular momentum.
\[{{\mu }_{B}}=\dfrac{e\hbar }{2{{m}_{e}}}\], where ‘e’ is the charge of an electron, ħ is the ratio of planck's constant by 2п, me is the mass of the electron.
It is basically a constant or sometimes chosen as a natural constant to express the magnetic moment caused by an electron. It is denoted by the simple B. Its value is $9.27400968 \times 10^{-24} JT^{-1}$ in SI units.
Nuclear magneton is the analogue of Bohr Magneton in case of heavier nucleons like protons and atomic nuclei.
\[{{\mu }_{N}}=\dfrac{e\hbar }{2{{m}_{p}}}\], where ‘e’ is the charge of an electron, ħ is the ratio of planck's constant by 2$\pi$, $m_p$ is the mass of proton.
Now we know what is nuclear magneton and Bohr magneton. So, let us find its ratio. The ratio of Bohr magneton to nuclear magneton is,
\[\dfrac{{{\mu }_{B}}}{{{\mu }_{N}}}=\dfrac{\dfrac{e\hbar }{2{{m}_{e}}}}{\dfrac{e\hbar }{2{{m}_{p}}}}=\dfrac{{{m}_{p}}}{{{m}_{e}}}\]
So, the correct option is option A.
Note: The things to be noted for further questions like this are:
1. Remember the formula of nuclear magneton and Bohr magneton. Their equations are different only in terms of mass. Bohr magneton uses electron mass while nuclear magneton uses proton mass. Be careful about it.
2. Understanding the concept of nuclear magneton and Bohr magneton is also good.
3. Don’t mix up with Nuclear magneton and Bohr magneton. Bohr magneton is for electron and nuclear magneton is for protons and other heavier nucleons. This is a common mistake.
Formula used:
\[{{\mu }_{B}}=\dfrac{e\hbar }{2{{m}_{e}}}\]
\[{{\mu }_{N}}=\dfrac{e\hbar }{2{{m}_{p}}}\]
Complete step by step solution:
Bohr Magneton is the magnetic moment of an electron caused by its spin or angular momentum.
\[{{\mu }_{B}}=\dfrac{e\hbar }{2{{m}_{e}}}\], where ‘e’ is the charge of an electron, ħ is the ratio of planck's constant by 2п, me is the mass of the electron.
It is basically a constant or sometimes chosen as a natural constant to express the magnetic moment caused by an electron. It is denoted by the simple B. Its value is $9.27400968 \times 10^{-24} JT^{-1}$ in SI units.
Nuclear magneton is the analogue of Bohr Magneton in case of heavier nucleons like protons and atomic nuclei.
\[{{\mu }_{N}}=\dfrac{e\hbar }{2{{m}_{p}}}\], where ‘e’ is the charge of an electron, ħ is the ratio of planck's constant by 2$\pi$, $m_p$ is the mass of proton.
Now we know what is nuclear magneton and Bohr magneton. So, let us find its ratio. The ratio of Bohr magneton to nuclear magneton is,
\[\dfrac{{{\mu }_{B}}}{{{\mu }_{N}}}=\dfrac{\dfrac{e\hbar }{2{{m}_{e}}}}{\dfrac{e\hbar }{2{{m}_{p}}}}=\dfrac{{{m}_{p}}}{{{m}_{e}}}\]
So, the correct option is option A.
Note: The things to be noted for further questions like this are:
1. Remember the formula of nuclear magneton and Bohr magneton. Their equations are different only in terms of mass. Bohr magneton uses electron mass while nuclear magneton uses proton mass. Be careful about it.
2. Understanding the concept of nuclear magneton and Bohr magneton is also good.
3. Don’t mix up with Nuclear magneton and Bohr magneton. Bohr magneton is for electron and nuclear magneton is for protons and other heavier nucleons. This is a common mistake.
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