Question

Magnetic field at the center (at nucleus) of the hydrogen-like atoms (atomic number $=Z$) due to the motion of electron in $n^{th}$ orbit is proportional to

• A. $\frac{n^3}{z^5}$
• B. $\frac{n^4}{z}$
• C. $\frac{z^2}{n^3}$
• D. $\frac{z^3}{n^5}$

Answer 1

The correct option is D $\frac{z^3}{n^5}$

$m_e{\omega }_n{r}_n^2=\frac{nh}{2\pi }$ ................(i)

$F=m_e{a}_n=\frac{K{q}_e^2}{r_n^2}$

$a_n=\frac{K{q}_e^2}{m_e{r}_n^2}={\omega }_n^2{r}_n$

${\omega }_n^2{r}_n=\frac{q_e^2}{4\pi {ϵ}_o{m}_e{r}_n^2}$...............(ii)

Solving (i) and (ii) we get:

${\omega }_n=\frac{m_e{q}_e^4\pi }{2n^3{h}^3{ϵ}_o^2}$

$r_n=\frac{n^2{h}^2{ϵ}_o}{q_e^2{m}_e\pi }$

$I=q_e\frac{{\omega }_n}{2\pi }=\frac{q_e^5{m}_e}{4n^3{h}^3{ϵ}_o^2}$

$B={\mu }_o\frac{I}{2r_n}$

$B=\frac{{\mu }_o{q}_e^7{m}_e^2\pi }{8n^5{h}^5{ϵ}_o^3}$