Question

Magnet field at the center (at nucleus) of the hydrogen-like atoms (atomic number $ = {\text{Z}}$ ) due to the motion of electron in ${{\text{n}}^{{\text{th}}}}$ orbit is proportional to
A. $\dfrac{{{{\text{n}}^3}}}{{{{\text{z}}^5}}}$
B. $\dfrac{{{{\text{n}}^4}}}{{\text{z}}}$
C. $\dfrac{{{{\text{z}}^2}}}{{{{\text{n}}^3}}}$
D. $\dfrac{{{{\text{z}}^3}}}{{{{\text{n}}^5}}}$

Answer 1

Hint:In this question analyse angular momentum equation. Solve force equation that is ${\text{F = }}{{\text{m}}_{\text{e}}}{{\text{a}}_{\text{n}}}$ and the equation given by the coulomb law equation. After that write the current equation with the amount of charge flowing in given time and then magnetic field equation to get the answer.

Complete step by step answer:
Angular momentum equation of the given
${{\text{m}}_{\text{e}}}{\omega _{\text{n}}}{{\text{r}}_{\text{n}}}^{\text{2}} = \dfrac{{{\text{nh}}}}{{2\pi }}\,\,\,.......(i)$
Force equation:
${\text{F}}\,{\text{ = }}\,{{\text{m}}_{\text{e}}}{{\text{a}}_{\text{n}}}\, = \,\dfrac{{{\text{K}}{{\text{q}}_{\text{e}}}^2}}{{{{\text{r}}_{\text{n}}}^{\text{2}}}}$
$ \Rightarrow {{\text{a}}_{\text{n}}}\, = \,\dfrac{{{\text{K}}{{\text{q}}_{\text{e}}}^2}}{{{{\text{m}}_{\text{e}}}{{\text{r}}_{\text{n}}}^{\text{2}}}}\, = \,{\omega _{\text{n}}}^2{{\text{r}}_{\text{n}}}$

Now put ${\text{K}}\,{\text{ = }}\,\dfrac{1}{{{\text{4}}\pi {\varepsilon _0}}}$ in above equation
${\omega _{\text{n}}}^2{{\text{r}}_{\text{n}}}\, = \,\dfrac{{{{\text{q}}_{\text{e}}}^2}}{{{\text{4}}\pi {\varepsilon _0}{{\text{m}}_{\text{e}}}{{\text{r}}_{\text{n}}}^{\text{2}}}}\,\,\,\,\,......(ii)$
Solving $(i)$ and $(ii)$ we get:
${\omega _{\text{n}}}\, = \,\dfrac{{{{\text{m}}_{\text{e}}}{{\text{q}}_{\text{e}}}^4\pi }}{{2{{\text{n}}^3}{{\text{h}}^3}{\varepsilon _0}^2}}$
and ${{\text{r}}_{\text{n}}}\, = \,\dfrac{{{{\text{n}}^2}{{\text{h}}^2}{\varepsilon _0}}}{{{{\text{q}}_{\text{e}}}^2{{\text{m}}_{\text{e}}}\pi }}$

Now write the relation of current, charge and time:
${\text{I}}\,{\text{ = }}\,{{\text{q}}_{\text{e}}}\dfrac{{{\omega _{\text{n}}}}}{{2\pi }}\, = \,\,\dfrac{{{{\text{m}}_{\text{e}}}{{\text{q}}_{\text{e}}}^5}}{{{\text{4}}{{\text{n}}^3}{{\text{h}}^3}{\varepsilon _0}^2}}$
By writing the equation of magnetic field we get:
${\text{B = }}{\mu _0}\dfrac{{\text{I}}}{{2{{\text{r}}_{\text{n}}}}}$
$ \therefore {\text{B = }}\dfrac{{{\mu _0}{{\text{m}}_{\text{e}}}^2{{\text{q}}_{\text{e}}}^7\pi }}{{{\text{8}}{{\text{n}}^5}{{\text{h}}^5}{\varepsilon _0}^3}}\,\,\,\,.....(iii)$

Hence, the correct option is D.

Note:Motion of electrons around the hydrogen atom produces a magnetic field at the center of the atom. Due to the continuous movement of the charges magnetic field is generated which we can calculate as above. This magnetic field value depends on the electron’s orbital number.