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An Electron in an Atom Revolves Around the Nucleus in an Orbit of Radius 0.53 Å. If the Frequency of Revolution of an Electron - Physics

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Question

An electron in an atom revolves around the nucleus in an orbit of radius 0.53 Å. If the frequency of revolution of an electron is 9 x109 MHz, calculate the orbital  angular momentum

[Given : Charge on an electron = 1.6 x 10–19 C; Gyromagnetic ratio = 8.8 x 1010 C/kg; π = 3.142]

Solution

Given : Charge on an electron = 1.6 x 10–19 C;
Gyromagnetic ratio = 8.8 x 1010 C/kg; π = 3.142

Formula:  `L_0=M_0/"gyromagnetic ratio"`

M=IA

Since `I=1/T e=fe`

From formula 

`M=feA=fepir^2`

`=9xx10^15xx1.6xx10^-19xxpixx(0.53xx10^-10)^2`

`=1.6xxpixx0.25xx10^-23`

`M=1.270xx10^-23 Am^2`

Using formula `L_0=(1.270xx10^-23)/(8.8 xx10^10)`

`L_0=0.1443xx10^-33 (kgm^2)/s`

  Is there an error in this question or solution?
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APPEARS IN

 2016-2017 (March) (with solutions)
Question 7.4 | 3.00 marks
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An Electron in an Atom Revolves Around the Nucleus in an Orbit of Radius 0.53 Å. If the Frequency of Revolution of an Electron Concept: Magnetic Dipole Moment of a Revolving Electron.
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