An electron revolving in the n^{text{th}} Boh | vedclass

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(B) The magnetic moment $\mu$ of an electron revolving in an orbit is given by $\mu = iA$,where $i$ is the current and $A$ is the area of the orbit.$i

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$1$

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(B) The magnetic moment $\mu$ of an electron revolving in an orbit is given by $\mu = iA$,where $i$ is the current and $A$ is the area of the orbit.$i = \frac{e}{T} = \frac{ev}{2\pi r}$,where $v$ is the velocity and $r$ is the radius.$A = \pi r^2$.Therefore,$\mu = \left(\frac{ev}{2\pi r}\right) \pi r^2 = \frac{evr}{2}$.According to Bohr's theory,$r \propto n^2$ and $v \propto \frac{1}{n}$.Substituting these relations: $\mu \propto \left(\frac{1}{n}\right) \cdot n^2 = n^1$.Comparing $\mu \propto n^1$ with $\mu \propto n^x$,we get $x = 1$.

An electron revolving in the $n^{\text{th}}$ Bohr orbit has a magnetic moment $\mu_n$. If $\mu_n \propto n^x$,the value of $x$ is:

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