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In the Bohr model an electron moves in a circular orbit around the proton. Considering the orbiting electron to be a circular current loop, the magnetic moment of the hydrogen atom, when the electron is in nth excited state is :

Option: 1

(\frac{e}{2m})\frac{n^{2}h}{2\pi }
 


Option: 2

(\frac{e}{m})\frac{nh}{2\pi }


Option: 3

(\frac{e}{2m})\frac{nh}{2\pi }

 


Option: 4

(\frac{\Theta }{m})\frac{n^{2}h}{2\pi }


Answers (1)

best_answer

As , i=\frac{e}{T}

 and magnetic moment  

\\ M=i A \quad\left(\because A=\pi r^{2}\right)\\ \\ \therefore M=\frac{e}{T} \cdot \pi r^{2}

\begin{aligned} &T=\frac{2 \pi r}{v}\\ &\text { It becomes, } M=\frac{\frac{e}{2 \pi r} \cdot \pi r^{2}}{v}=\frac{e v r}{2} \quad \text { (ii) } \\ \\ \quad \text { Also, } \quad \text { mvr }=\frac{n h}{2 \pi} \end{aligned}

\begin{aligned} &v r=\frac{n h}{2 \pi m}\\ &\text { Putting this value in Eq. (ii), we get } \quad M=\frac{e \cdot n h}{2.2 \pi m}\\ \\ =\left(\frac{e}{2 m}\right) \frac{n h}{2 \pi} \end{aligned}

Posted by

Divya Prakash Singh

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