Question

The ratio of nuclear magneton and Bohr magneton is
A. ${{m}_{e}}/{{m}_{p}}$
B. ${{m}_{p}}/{{m}_{e}}$
C. ${{m}_{e}}{{m}_{p}}$
D. $2{{m}_{p}}/{{m}_{ez}}$

Answer 1

Magneton refers to a unit of magnetic moment. It is calculated by multiplying the magnet’s pole strength to the distance between its poles. For Bohr magneton, mass of an electron is taken into account while for the nuclear magneton, a proton’s mass is taken into account
Bohr magneton = $µ_B$ = 9.274 $\times 10^{-21} JT^{-1}$
Nuclear magneton = $µ_n$ = 5.051 $\times 10^{-27} JT^{-1}$

Formula used:
${{\mu }_{B}}=\dfrac{eh}{2{{m}_{e}}}$ (Bohr’s Magneton)
${{\mu }_{n}}=\dfrac{eh}{2{{m}_{p}}}$ (Nuclear magneton)

Complete step-by-step answer:
As we know,
${{\mu }_{B}}=\dfrac{eh}{2{{m}_{e}}}$
And,
${{\mu }_{n}}=\dfrac{eh}{2{{m}_{p}}}$
Where, $µ_B$ and $µ_n$ are Bohr’s magneton and nuclear magneton respectively
So, the ratio of nuclear magneton and Bohr magneton will be
$\dfrac{{{\mu }_{n}}}{{{\mu }_{B}}}=\dfrac{\dfrac{eh}{2{{m}_{p}}}}{\dfrac{eh}{2{{m}_{e}}}}$
$\dfrac{{{\mu }_{n}}}{{{\mu }_{B}}}=\dfrac{eh}{2{{m}_{p}}}\times \dfrac{2{{m}_{e}}}{eh}$
$\dfrac{{{\mu }_{n}}}{{{\mu }_{B}}}=\dfrac{{{m}_{e}}}{{{m}_{p}}}$

So, the ratio between nuclear magneton and Bohr’s magneton will be Option A ${{m}_{e}}/{{m}_{p}}$

So, the correct answer is “Option A”.

Additional Information: Bohr’s magneton is named after the 20th century, Danish physicist Niels Bohr who won the 1922 Nobel Prize in physics for his commendable work on investigation on the structure of atoms. He was also the student of another Nobel prize winner Ernest Rutherford, who did his research on the same topic.

Note: Niels Bohr is widely known as the father of atomic physics because of his contribution in the atomic model of an atom. During the World War II, Niels Bohr was forced flee from Copenhagen the capital of Denmark to save himself from Nazi prosecution that was led by Hitler. After the war he returned to Denmark and advocated for the peaceful use of the nuclear weapons.