Assuming the atom is in the ground state, the | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

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

Vedclass · Answer

(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}$.

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

Excitation energy of a hydrogen-like ion in its first excitation state is $40.8 \, eV$. Energy needed to remove the electron from the ion in the ground state is ........ $eV$.

If $E$ and $L$ denote the magnitudes of total energy and of angular momentum of an electron in a Bohr orbit,then the true relation between them is:

Which of the following statements about the Bohr model of the hydrogen atom is false?

An electron rotates in a circle around a nucleus having positive charge $Ze$. The correct relation between the total energy $(E)$ of the electron and its potential energy $(U)$ is:

The radius of the first orbit of a hydrogen atom is $0.5 \, \mathring{A}$ and the speed of the electron is $2.2 \times 10^6 \, m/s$. Find the magnetic induction at the proton due to the motion of the electron in $Tesla$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module