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(A) The magnetic field $B$ at the center of a circular current loop is given by $B = \frac{\mu_0 I}{2r}$.Since the electron revolves with period $T$,

Vedclass · Accepted Answer

$\frac{\mu_0 e^7 \pi m^2}{8 \epsilon_0^3 h^5}$

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(A) The magnetic field $B$ at the center of a circular current loop is given by $B = \frac{\mu_0 I}{2r}$.Since the electron revolves with period $T$, the equivalent current is $I = \frac{e}{T}$.Given $T = \frac{2\pi r}{v}$, we have $I = \frac{ev}{2\pi r}$.Substituting this into the magnetic field formula: $B = \frac{\mu_0 (ev/2\pi r)}{2r} = \frac{\mu_0 ev}{4\pi r^2}$.For a hydrogen atom in the ground state $(n=1)$, the velocity $v$ and radius $r$ are given by Bohr's theory:$v = \frac{e^2}{2\epsilon_0 h}$ and $r = \frac{\epsilon_0 h^2}{\pi m e^2}$.Substituting these into the expression for $B$:$B = \frac{\mu_0 e}{4\pi} \cdot \left( \frac{e^2}{2\epsilon_0 h} \right) \cdot \left( \frac{\pi m e^2}{\epsilon_0 h^2} \right)^2$$B = \frac{\mu_0 e^3}{8\pi \epsilon_0 h} \cdot \frac{\pi^2 m^2 e^4}{\epsilon_0^2 h^4} = \frac{\mu_0 e^7 \pi m^2}{8 \epsilon_0^3 h^5}$.

$A$. Speed of revolution of electron	$i$. $\frac{1}{4 \pi \varepsilon_0} \frac{2 \pi Z e^2}{n h}$
$B$. Kinetic energy	$ii$. $-\left(\frac{1}{4 \pi \varepsilon_0}\right)^2 \frac{2 \pi^2 m e^4 Z^2}{n^2 h^2}$
$C$. Total energy	$iii$. $\left(\frac{1}{4 \pi \varepsilon_0}\right)^2 \frac{2 \pi^2 m e^4 Z^2}{n^2 h^2}$
$D$. Frequency	$iv$. $\left(\frac{1}{4 \pi \varepsilon_0}\right)^2 \frac{4 \pi^2 Z^2 e^4 m}{n^3 h^3}$

Assuming the atom is in the ground state, the expression for the magnetic field at the nucleus in a hydrogen atom due to the circular motion of the electron is [$\mu_0 =$ permeability of free space, $m =$ mass of electron, $\epsilon_0 =$ permittivity of free space, $h =$ Planck's constant].

Similar Questions

According to de-Broglie,the de-Broglie wavelength for an electron in an orbit of radius $5.3 \times 10^{-11} \ m$ of a hydrogen atom is $10^{-10} \ m$. The principal quantum number for this electron is:

Angular momentum of an electron in a hydrogen atom is $\frac{3h}{\pi}$, then the energy of the electron is . . . . . . eV.

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