An electron revolving in the \( n^{th} \) Bohr orbit has magnetic moment \( \mu \). If \( \mu_n \) is the value of \( \mu \), the value of \( x \) is:
An electron revolving in the \( n^{th} \) Bohr orbit has magnetic moment \( \mu \). If \( \mu_n \) is the value of \( \mu \), the value of \( x \) is:
2
1
3
0
The Correct Option is B
Solution and Explanation
The magnetic moment \( \mu \) of an electron in the \( n^{th} \) Bohr orbit is given by the formula:
\[ \mu = \frac{e}{2m} r^2, \]
where:
- \( e \) is the charge of the electron,
- \( m \) is the mass of the electron,
- \( r \) is the radius of the orbit.
For a hydrogen atom, the radius of the \( n^{th} \) Bohr orbit is given by:
\[ r_n = n^2 \frac{h^2}{4\pi^2 ke^2m}, \]
where \( h \) is Planck’s constant and \( k \) is Coulomb’s constant.
Substituting \( r_n \) into the magnetic moment formula gives:
\[ \mu_n = \frac{e}{2m} \left( n^2 \frac{h^2}{4\pi^2 ke^2m} \right) = \frac{eh^2n^2}{8\pi^2 ke^2m^2}. \]
This simplifies to:
\[ \mu_n = n^2 \left( \frac{eh^2}{8\pi^2 ke^2m^2} \right). \]
Since \( \mu_1 \) (the magnetic moment for the first orbit) can be taken as a reference, we can find the ratio:
\[ \frac{\mu_n}{\mu_1} = n^2. \]
Thus, when \( n = 1 \):
\[ \frac{\mu_n}{\mu_1} = 1^2 = 1. \]
Therefore, the value of \( x \) is: 1
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