Introduction :
Bohr assumed that an electron is a particle and postulated that it revolves around the nucleus in an orbit in which the angular momentum ( mvr ) of the electron is an integral multiple of `(h)/(2pi)` .
m v r = n`(h)/(2pi)` .
But according to de-Broglie an electron in motion with high speed is associated with wave nature. For an electron moving around the nucleus in the circular path, two cases of electron waves of different wavelengths are possible .
In one case the circumference of the electron orbit ( 2`pi`r ) is an integral multiple of the wavelength ( `lambda` ) of the electron. This is shown in the figure (1). In the second case, the circumference of the electron orbit is not an integral multiple of the wavelength . This is shown in figure (2) .
In the first case of atomic model above, the two ends of the electron wave meet to give a regular series of crests and troughs. In this case the electron waves are said to be in-phase. This means that there is constructive interference of the electron - waves. In such a case, the electron is considered to behave as a standing wave or non-energy radiating wave. This extends round the nucleus in a circular orbit.
In the second case of atomic model above, here the circumference of the Bohr's orbit ( 2`pi`r ) is bigger or smaller than n`lambda` , the electron waves are said to be out of phase. This is shown in the figure (2). In this case, a destructive interference of waves occur causing radiation of energy.
Therefore the necessary condition to get an electron - wave in phase is that the circumference of the Bohr's orbit ( 2`pi`r ) is equal to integral ( whole number ) multiple of the wave length (`lambda`) of the electron - wave
n`lambda` = 2`pi`r
`lambda` = `(2pir)/(n)` ; n = integer or whole number
According to de Broglie theory `lambda` = `(h)/(mv)`
`:.` `(2pir)/(n)` = `(h)/(mv)`
or m v r = n * `h/(2pi)`
This is Bohr's postulate that stipulates the angular momentum of an electron moving round the nucleus is an integral multiple of `h/(2pi)` . In other words, de Broglie theory and the Bohr's theory are in agreement with each other. In case the circumference of the Bohr's orbit (2`pi`) is bigger or smaller than n`lambda` , destructive interference of waves takes place and the orbit cannot exist.
Solution : de Broglie Wavelength , `lambda` = 1 A = 1 * 10-10 m
As per de Broglie's equation `lambda` = `h/p`
`:.` Momentum of particle , p = `h/lambda`
= `(6.6256*10^-34kgm^2s^-1)/(1*10^-10m)`
Momentum of particle, p = 6.6256*10-24 kgms-1 .
Solution : Given mass of the ball m = 8 kg and velocity v = 9 m / s
a ) de Broglie's Wavelength is given by `lambda` = `h/p`
where p = m v and h is Planck's constant .
In this case `lambda` = `(6.6256*10^-34)/(8*9)`
`lambda` = 0.092 * 10-34
b ) Wave cannot be seen for macroscopic objects like bowling balls. The de Broglie wavelength for this object is smaller than the atomic nucleus and impossible to detect .
Bohr assumed that an electron is a particle and postulated that it revolves around the nucleus in an orbit in which the angular momentum ( mvr ) of the electron is an integral multiple of `(h)/(2pi)` .
m v r = n`(h)/(2pi)` .
But according to de-Broglie an electron in motion with high speed is associated with wave nature. For an electron moving around the nucleus in the circular path, two cases of electron waves of different wavelengths are possible .
In one case the circumference of the electron orbit ( 2`pi`r ) is an integral multiple of the wavelength ( `lambda` ) of the electron. This is shown in the figure (1). In the second case, the circumference of the electron orbit is not an integral multiple of the wavelength . This is shown in figure (2) .
In the first case of atomic model above, the two ends of the electron wave meet to give a regular series of crests and troughs. In this case the electron waves are said to be in-phase. This means that there is constructive interference of the electron - waves. In such a case, the electron is considered to behave as a standing wave or non-energy radiating wave. This extends round the nucleus in a circular orbit.
In the second case of atomic model above, here the circumference of the Bohr's orbit ( 2`pi`r ) is bigger or smaller than n`lambda` , the electron waves are said to be out of phase. This is shown in the figure (2). In this case, a destructive interference of waves occur causing radiation of energy.
Therefore the necessary condition to get an electron - wave in phase is that the circumference of the Bohr's orbit ( 2`pi`r ) is equal to integral ( whole number ) multiple of the wave length (`lambda`) of the electron - wave
n`lambda` = 2`pi`r
`lambda` = `(2pir)/(n)` ; n = integer or whole number
According to de Broglie theory `lambda` = `(h)/(mv)`
`:.` `(2pir)/(n)` = `(h)/(mv)`
or m v r = n * `h/(2pi)`
This is Bohr's postulate that stipulates the angular momentum of an electron moving round the nucleus is an integral multiple of `h/(2pi)` . In other words, de Broglie theory and the Bohr's theory are in agreement with each other. In case the circumference of the Bohr's orbit (2`pi`) is bigger or smaller than n`lambda` , destructive interference of waves takes place and the orbit cannot exist.
Solved Problem on De broglie atomic model:
Q ) Find the momentum of a particle whose de Broglie wavelength is 1A.Solution : de Broglie Wavelength , `lambda` = 1 A = 1 * 10-10 m
As per de Broglie's equation `lambda` = `h/p`
`:.` Momentum of particle , p = `h/lambda`
= `(6.6256*10^-34kgm^2s^-1)/(1*10^-10m)`
Momentum of particle, p = 6.6256*10-24 kgms-1 .
Another problem on De broglie atomic model:
Q ) A 8 kg bowling ball rolls with a velocity of 9 metre per second. a ) What is de Broglie's wavelength of the bowling ball. b) Why does the bowling ball exhibit no observable wave behaviour ?Solution : Given mass of the ball m = 8 kg and velocity v = 9 m / s
a ) de Broglie's Wavelength is given by `lambda` = `h/p`
where p = m v and h is Planck's constant .
In this case `lambda` = `(6.6256*10^-34)/(8*9)`
`lambda` = 0.092 * 10-34
b ) Wave cannot be seen for macroscopic objects like bowling balls. The de Broglie wavelength for this object is smaller than the atomic nucleus and impossible to detect .
No comments:
Post a Comment