Name: Bohr Atom Model: Frequency of Emitted Radiation
Uploaded: 2019-07-19
Duration: 3 min 54 s
Description: Electrons jump in different energy levels. Let's understand how we can calculate the frequency of emitted and absorbed radiation as electrons transition into...

Question

An electron of charge $e$ is revolving around a nucleus along a circular path of the radius $r$ with speed $v$ . Starting from the relation $μ_{l} = IA$ , where the symbols have their usual meaning, show that,

(1) $μ_{l} = \frac{evr}{2}$ and (2) $μ_{l} = - \frac{e \vec{l}}{2 m_{e}}$ .

Here $l$ = magnitude of angular momentum of the electron of mass $m_{e}$

Accepted Answer

Given $μ_{l} = IA$

where $I$ is the current and $A$ is the area of the loop. The current associated with an electron in orbit about a proton in a hydrogen atom is

$I = \frac{e}{T}$ and $T = \frac{path}{speed} = \frac{2 πr}{v}$
where $e$ is the magnitude of the electron charge and $T$ is its orbital period. $r$ is the radius of the orbit and $v$ is the speed of the electron in its orbit. Given that the area of a circle is ${πr}^{2}$ , the absolute magnetic moment is

$μ_{l} = IA = \frac{e}{(\frac{2 πr}{v})} . {πr}^{2} = \frac{evr}{2}$ ...................................................(A)

As the angular momentum $l$ of the electron is given by the relation.

$l = r \times p = r m_{e} v or r . v = \frac{l}{m_{e}}$ , where $m_{e}$ = mass of the electron.

Substituting this value in equation (A), we get,

$μ_{l} = \frac{e . l}{2 m_{e}}$ ..................................................(B)

Important Questions on Atoms

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An electron of charge e is revolving around a nucleus along a circular path of the radius r with speed v. Starting from the relation μl=IA, where the symbols have their usual meaning, show that, (1) μl=evr2 and (2)μl=-el→2 me. Here l= magnitude of angular momentum of the electron of mass me

Important Questions on Atoms

Learn and Prepare for any exam you want !