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प्रश्न
Obtain Bohr’s quantisation condition for angular momentum of electron orbiting in nth orbit in hydrogen atom on the basis of the wave picture of an electron using de Broglie hypothesis.
उत्तर १
`n lambda = 2 pi r`
But ` lambda = n/(mv) `( By De Broglie)
`(n lambda) /(mv) = 2 pi r`
` therefore (lambda n) /(2 pi ) = mvr ` (mvr = L)
` therefore L = (lambda n)/(2 pi)`
उत्तर २
Bohr's second postulate states that the angular momentum of an electron has only those values that are integral multiples of `"h"/(2pi)` He thought that the motion of electrons within an atom is associated with the standing wave along the orbit as shown.
About standing waves in stretched strings, we know that only those waves survive for which the distances traveled in the round trip between the ends are integral multiples of the wavelength. Similarly, for an electron moving in the nth orbit of radius rn, the distance traveled in one trip is 2πrn, which should be an integral multiple of the wavelength.
2πrn = nλ (where n= 1, 2, 3, 4 etc)
By de-Broglie hypothesis, we have:
`lambda = "h"/"p" = "h"/("mv"_"n")`
Substituting the value of λ in the above expression, we get:
`2pi"r"_"n" = "n" "h"/"mv"_"n"`
⇒ `"mv"_"n""r"_"n" = "n" "h"/(2pi)`
Angular momentum `= "L" = "n" "h"/(2pi)`
संबंधित प्रश्न
What is the maximum number of emission lines when the excited electron of an H atom in n = 6 drops to the ground state?
The ratio of kinetic energy of an electron in Bohr’s orbit to its total energy in the same orbit is
(A) – 1
(B) 2
(C) 1/2
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