Assuming the atom is in the ground state, the expression for the magnetic field at a point nucleus in…

Assuming the atom is in the ground state, the expression for the magnetic field at a point nucleus in hydrogen atom due to circular motion of electron is $\left[\mu_0=\right.$ permeability of free space, $m$ $=$ mass of electron, $\in_0=$ permittivity of free space, $\mathrm{h}=$ Planck's constant]
  1. $\frac{\mu_0 \mathrm{e}^7 \pi \mathrm{m}^2}{8 \epsilon_0^3 \mathrm{~h}^5}$
  2. $\frac{\mu_0 \mathrm{e}^5 \pi^2 \mathrm{~m}^2}{8 \epsilon_0^2 \mathrm{~h}^4}$
  3. $\frac{\mu_0 \mathrm{e}^5 \pi \mathrm{m}^3}{8 \epsilon_0^3 \mathrm{~h}^5}$
  4. $\frac{\mu_0 \mathrm{e}^7 \pi^2 \mathrm{~m}^2}{8 \epsilon_0^3 \mathrm{~h}^5}$

Solution

The magnetic field at the center of a circular coil is given by $\mathrm{B}=\frac{\mu_0 \mathrm{I}}{2 \mathrm{r}}$ If $\mathrm{T}$ is the periodic time of revolving electron, then $\mathrm{I}=\frac{\mathrm{e}}{\mathrm{T}}$ Also, $\mathrm{T}=\frac{2 \pi \mathrm{r}}{\mathrm{V}}$ $\therefore \mathrm{I}=\frac{\mathrm{eV}}{2 \pi \mathrm{r}}$ $\therefore \mathrm{B}=\frac{\mu_0 \mathrm{eV}}{4 \pi \mathrm{r}^2}$ For electron in ground state, $\mathrm{V}=\frac{\mathrm{e}^2}{2 \in_0 \mathrm{~h}} \text { and } \mathrm{r}=\frac{\mathrm{h}^2 \in_0}{\pi \mathrm{m}^2}$ Putting these values in eq.(1) and simplifying we get $\therefore \mathrm{B}=\frac{\mu_0 \mathrm{e}^7 \pi \mathrm{m}^2}{8 \epsilon_0^3 \mathrm{~h}^5}$

Asked in: MHT CET 2021 (22 Sep Shift 1)

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