Question

Magnetic field at the center (at nucleus) of the hydrogen-like atoms (atomic number = Z) due to the motion of an electron in nth orbit is proportional to

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

Answer 1

The correct option is D

$\frac{{{\mathrm{z}}^3}^{}}{{{\mathrm{n}}^5}^{}}$

Explanation for the correct answer:-

Option (D)

Step 1: Magnetic Field Induction formula

$\mathrm{B}=\frac{{\mathrm{\mu }}_0\mathrm{i}}{2\mathrm{R}}$

where 'i' is current

'R' is the radius of an orbit

'${\mathrm{\mu }}_0$' is the permeability of free space

or $\mathrm{B}\propto \frac{\mathrm{i}}{\mathrm{R}}$

and $\mathrm{i}=\left(\frac{\mathrm{q}}{\mathrm{t}}\right)$

where ‘q’ is the quantity of charge and 't' is time

Therefore, $\mathrm{B}\propto \frac{\mathrm{q}}{\mathrm{tR}}$

Now, $\mathrm{R}\propto \frac{{\mathrm{n}}^2}{\mathrm{z}}\left[\because \mathrm{R}=\frac{{\mathrm{n}}^2{\mathrm{h}}^2{\in }_0}{{\mathrm{\pi me}}^2\mathrm{z}}\right]$

Therefore, $\mathrm{B}\propto \frac{\mathrm{qz}}{{\mathrm{tn}}^2}$

Step 2: Finding the magnetic field induction

Since, $\mathrm{t}=\frac{2\mathrm{\pi r}}{\mathrm{v}}$

$\mathrm{v}\propto \frac{\mathrm{z}}{\mathrm{n}}\mathrm{and}\mathrm{r}\propto \frac{{\mathrm{n}}^2}{\mathrm{z}}$

$\mathrm{B}\propto \frac{\mathrm{qz}{\displaystyle \frac{\mathrm{z}}{\mathrm{n}}}}{{\displaystyle \frac{{\mathrm{n}}^2}{\mathrm{z}}}{\mathrm{n}}^2}$

$$\begin{gathered} \Rightarrow \mathrm{B}\propto \frac{{\mathrm{qz}}^3}{{\mathrm{n}}^5} \\ \Rightarrow \mathrm{B}\propto \frac{{\mathrm{z}}^3}{{\mathrm{n}}^5} \end{gathered}$$

Hence it is the correct option.

Therefore option D is the correct answer.