Derivation of angular momentum from de broglie equation
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Derivation of Angular Momentum from de Broglie Equation :
According to Bohr’s model, the electron revolves around the nucleus in circular orbits. According to de Broglie concept, the electron is not only a particle but has a wave character also.
If the wave is completely in phase, the circumference of the orbit must be equal to an integral multiple of wave length (l)
Therefore 2πr = nλ
where ‘ n ’ is an integer and ‘ r ’ is the radius of the orbit
But l = h/mv
∴ 2πr = nh /mv
(or) mvr = n h/2π
which is Bohr’s postulate of angular momentum, where ‘n’ is the principal
quantum number.
Thus, the number of waves an electron makes in a particular Bohr orbit in one complete revolution is equal to the principal quantum number of the orbit ”.
Alternatively
Number of waves ‘ n ’ = 2πr /λ
where v and r are the velocity of electron and radius of that particular Bohr orbit in which number of waves are to be calculated, respectively.
The electron is revolving around the nucleus in a circular orbit. How many revolutions it can make in one second
Let the velocity of electron be v m/sec. The distance it has to travel for one revolution 2πr , (i.e., the circumference of the circle).
Thus, the number of revolutions per second is = v/2πr
Common unit of energy is electron volt which is amount of energy given when an electron is accelerated by a potential of exactly 1 volt. This energy equals the product of voltage and charge.
Since in SI units coulombs × volts = Joules, 1 eV numerically equals the electronic charge except that joule replaces coulombs.
Hope it helps...
According to Bohr’s model, the electron revolves around the nucleus in circular orbits. According to de Broglie concept, the electron is not only a particle but has a wave character also.
If the wave is completely in phase, the circumference of the orbit must be equal to an integral multiple of wave length (l)
Therefore 2πr = nλ
where ‘ n ’ is an integer and ‘ r ’ is the radius of the orbit
But l = h/mv
∴ 2πr = nh /mv
(or) mvr = n h/2π
which is Bohr’s postulate of angular momentum, where ‘n’ is the principal
quantum number.
Thus, the number of waves an electron makes in a particular Bohr orbit in one complete revolution is equal to the principal quantum number of the orbit ”.
Alternatively
Number of waves ‘ n ’ = 2πr /λ
where v and r are the velocity of electron and radius of that particular Bohr orbit in which number of waves are to be calculated, respectively.
The electron is revolving around the nucleus in a circular orbit. How many revolutions it can make in one second
Let the velocity of electron be v m/sec. The distance it has to travel for one revolution 2πr , (i.e., the circumference of the circle).
Thus, the number of revolutions per second is = v/2πr
Common unit of energy is electron volt which is amount of energy given when an electron is accelerated by a potential of exactly 1 volt. This energy equals the product of voltage and charge.
Since in SI units coulombs × volts = Joules, 1 eV numerically equals the electronic charge except that joule replaces coulombs.
Hope it helps...
Answered by
10
Let us consider an electron revolving around a nucleus of an atom. The wave train formed by it is as shown in figure. Now, for the wave to be in phase, the circumference of the orbit around the nucleus should be an integral multiple of the wavelength of the wave.
2Ωr=nλ,
where r is the radius of the orbit.
=nh/mν (from de Broglie’s equation)
Therefore,
Angular momentum; mνr=nh/2Ω, electron being revolving in nth orbit.
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