Question

magnetic field due to revolution of electron in a hydrogen atom in ground state at centre is b the magnetic field at centre of first exited state will be.
​

Answer 1

MAGNETIC FIELD INTENSITY DUE TO ELECTRON REVOLUTION IN ORBIT:

• Movement of electron in a circular orbit can be considered as equivalent to a current carrying circular ring.

• Here, current will be i = q/t = e/T.

$B = \dfrac{ \mu_{0}(i) }{4\pi r} \times 2\pi$

$\implies B = \dfrac{ \mu_{0}(i) }{2r}$

$\implies B = \dfrac{ \mu_{0} \times \dfrac{e}{T} }{2r}$

$\implies B = \dfrac{ \mu_{0}e}{2rT}$

$\implies B \propto \dfrac{1}{r}$

• Now, we know that radius of excited state in hydrogen is r_(0) × n² (n is the orbit)

$\implies B \propto \dfrac{1}{r_{0} \times {n}^{2} }$

$\implies B \propto \dfrac{1}{ {n}^{2} }$

So, at first excited state, when n = 2 , the Magnetic Field will be :

$B_{excited} = \dfrac{b}{ {n}^{2} }$

$\implies B_{excited} = \dfrac{b}{ {2}^{2} }$

$\implies B_{excited} = \dfrac{b}{ 4 }$

So, final answer is b/4.