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>> Derivation of Angular Momentum from de Broglie Equation
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 (λ)
Therefore 2πr = nλ
where ‘n’ is an integer and ‘r’ is the radius of the orbit
But λ = 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”.
Number of waves ‘n’ = 2πr / λ = 2πr / h/mv = 2πmvr / h
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 x volts = joules, 1 eV numerically equals the electronic charge except that joule replaces coulombs.