Qs. What are the basic postulates of Bohr’s Atomic Thoery?
Introduction
Neil Bohr studied the spectrum of hydrogen atom. On the basis of his study, he proposed a theory, which is known as Bohr’s Atomic theory.
POSTULATES OF BOHR’S ATOMIC THEORY
The important Postulates of Bohr’s Atomic Theory are as follows:
Angular Momentum
Electrons revolve only in those orbits for which its orbital angular momentum is an integral multiple of h/2π, i.e.
L = mvr(n) = nh / 2π
Where,
m = mass of electron
V = velocity of electron
r(n) = radius of nth orbit
n = Principal quantum number
h = Plank’s Constant
1. Energy
The total energy of an electron remains constant as long as it remains in the same orbit. i.e. it does not radiate energy while revolving around the nucleus.
2. Energy Release
When an electron jumps from a higher orbit having energy ‘En’ to a lower orbit having energy ‘Ep’ then energy is released in the form of energy ‘hv’ i.e.
Eo – Ep = hv = hc / λ
Where,
v = Frequency of Photon
λ = Wavelength of Photon
c = Speed of light
h = Plank’s constant
Qs. Find out the radius, energy and wave number of hydrogen atom with the help of Bohr’s Atomic theory.
HYDROGEN ATOM
A hydrogen atom is the simplest of all atoms. It consist of a proton in the nucleus and an electron revolving around the nucleus.
RADIUS OF HYDROGEN ORBIT
Consider an electron of charge ‘-e’ revolving in a hydrogen atom around a proton of charge ‘+e’ with constant speed v.
When the electron revolves around the nucleus, then two forces balance its motion.
Coulomb’s Force = F = ke² / r² ——– (I)
Centrifugal Force = F = mv² / r ——- (II)
Comparing eq (I) and (II)
ke² / r² = mv² / r
=> ke² / mv² = r² / r
=> r = ke² / mv² ——– (III)
According to Bohr’s theory, angular momentum is an integral multiple of h/2π
mvr = nh / 2π
=> v = nh / 2π mr
=> 1/v = 2π mr / nh
Taking square of both sides
=> 1/v² = 4π² m² r² / n²h²
Substituting the above value in eq (III)
r = Ke² / m = 4π² m² r² / n²h²
=> r / r² = 4π² m k e² / n²h²
=> 1 / r = 4π² m k e² / n²h²
=> r = n²h² / 4π²m k e²
We know that,
k = 1 /4π Єo => 1 / k = 4π Єo
=> r = n²h² / 4 π² m e² x 4π Єo
=> r = n²h² Єo / π e²
The above equation gives the radius of hydrogen atom.
Radii of Various Orbits
Radius of first orbit of hydrogen atom is calculated by substituting the following values in the equation of radius.
n = 1
h = 6.25 x 10(-34) J.sec
m = 9.1 x 10(-31) kg
k = 9 x 10(9) Nm²/col²
e = 1.6 x 10(-19) col
r = (1)² (6.625 x 10(-34)² / 4² (9.1 x 10(-31) (9 x 10(9)) (1.6 x 10-19)²
=> r = 0.53 x 10(-10)m
=> r1 = 0.53 Aº
For other orbits radius is given by
r2 = (2)² x 0.53 Aº
r3 = (3)² x 0.53 Aº
Similarly,
rn = n² x 0.53 Aº
ENERGY OF HYDROGEN ELECTRON
An electron revolving in the orbit of hydrogen atom possesses kinetic energy as well as Potential Energy. Therefore, total energy is given by
E = K.E + P.E —— (I)
Kinetic Energy
When an electron revolves in the orbit, then coulomb’s force is balanced by centrifugal force
ke²/r² = mv²/r
=> mv² = ke²/r
=> 1/2 mv² = ke²/2r
=> K.E = ke²/2r
Potential Energy
Potential energy is given by
P.E = F.dr
=> P.E = Ke² / r² dr
=> P.E = ke² 1 / r² dr
=> P.E = ke² |-1/r|
=> P.E = -ke² [1/r - 1/∞]
=> P.E = -ke² (1/r – 0)
=> P.E = -ke² / r
Total Energy
Substituting the values of K.E and P.E in eq (I)
E = ke² / 2 – ke² / r
=> E = k2² / 2r
Since,
r = n² h²/ 4π² m k e²
=> E = ke² / 2 4π² m k e² / n² h²
=> |E = 2π² m k² e² / n² h²|
The above equation gives the energy of the orbits of hydrogen atom. Negative sign shows that the electron is bound with the nucleus. When energy of the electron becomes positive, then electron will leave the nucleus.
WAVE NUMBER
When art electron jumps from higher orbit to inner orbit, then it radiates energy in the form of photons.
Qs. Explain the spectrum of hydrogen atom.
SPECTRUM OF HYDROGEN ATOM
When an electron jumps from a higher orbit to a lower orbit, it radiates energy which appears in the form of a spectral line. A set of such spectral lines is known as hydrogen spectrum. Hydrogen spectrum is the simplest one which consists of five series.
1. Layman Series
When an electron jumps from a higher orbit to the first orbit, Laymen Series (ultra violet region) is obtained.
The wavelength and wave number of Laymen Series can be calculated by
v = R(H) (1/1² – 1/n²)
Where n = 2, 3, 4, ……
2. Balmer Series
When an electron jumps from a higher orbit to the second orbit then Balmer Series (visible region) is obtained.
The wavelength and wave number of Balmer Series can be calculated by
v = R(H) (1/2² – 1/n²)
Where n = 3, 4, 5, ……
3. Paschen Series
When an electron jumps from a higher orbit to the third orbit then Paschen Series (infra red region) is obtained.
The wavelength and wave number of Paschen Series can be calculated by
v = R(H) (1/3² – 1/n²)
Where n = 4, 5, 6, ……
4. Bracket Series
When an electron jumps from a higher orbit to the fourth orbit then Bracket Series (infra red region) is obtained.
The wavelength and wave number of Bracket Series can be calculated by
v = R(H) (1/4² – 1/n²)
Where n = 5, 6, 7, ……
5. Pfund Series
When an electron jumps from a higher orbit to the fifth orbit then Pfund Series (infra red region) is obtained.
The wavelength and wave number of Pfund Series can be calculated by
v = R(H) (1/5² – 1/n²)
Where n = 6, 7, …..
Qs. Write a note on spectra of X-rays. Also write down the properties.
Introduction
X-Rays were discovered by W.K. Roentgen are also known as Roentgen rays. These rays of shorter wavelength, ranging from 0.1 nm to i nm. X-rays are produced if heavier atoms are bombarded by energetic electrons.
PRODUCTION OF X-RAYS
A Filament F and target T are produced in a vacuum chamber and voltage V is applied across the ends. Electrons are produced by heating the filament. These electrons are accelerated towards the metal by applying very high voltage (several thousands volts). When electrons hit the target, then X-rays are produced. There are two types of spectra obtained from this experiment.
1. A continuous spectrum of frequencies or X-rays Brems Strahlung.
2. Characteristics spectrum or a line spectrum of a limited number of fairly definite frequencies.
1. Continuous Spectra
When electrons hit the metal target, a continuous spectrum of frequencies of X-rays is emitted. The frequencies depend upon the accelerating voltage and are very nearly independent of the material of target.
Continuous spectrum is produced when electrons pass close to the atomic nuclei. The are deflected and slowed down due to which they lose their energy. The energy lost by decelerating electrons appears in the form of photon in the X-ray range. The process is represented as
Atoms + e(Fast) —–> Atom + e(Slow) + hv
2. Characteristic Spectra
In the heavy atoms, electrons are assumed to be arranged in concentric shells at increasing distance from the nucleus. The electrons of inner shell are much tightly bound as compared to the electrons of outer shells. Therefore, a large amount of energy is required to displace them Consequently photons of larger energy are emitted when atoms are stabilized. Thus the transition of inner shell electrons gives rise to high-energy spectra or Characteristic spectra. To obtain characteristic spectra, target metal of higher atomic number is used.
The process of emission of characteristic spectra takes place as follows. When a highly energetic incident electrons knocks an electron from the k-shell, a vacancy occurs in that shell. This vacancy is filled by the arrival of an electron from outside the k-shell, emitting excess amount of energy in the form of photon.
If the electrons jumps only one shell and returns with the emission of X-rays to Y shell, then X-rays are termed as ‘Yα’ X-rays. If the electron jumps two shells and returns with emission of X-rays to suppose ‘Y’ shell, then X-rays are termed as ‘Yβ’ rays and so on, where Y may be K, L, M, ……
Introduction
Neil Bohr studied the spectrum of hydrogen atom. On the basis of his study, he proposed a theory, which is known as Bohr’s Atomic theory.
POSTULATES OF BOHR’S ATOMIC THEORY
The important Postulates of Bohr’s Atomic Theory are as follows:
Angular Momentum
Electrons revolve only in those orbits for which its orbital angular momentum is an integral multiple of h/2π, i.e.
L = mvr(n) = nh / 2π
Where,
m = mass of electron
V = velocity of electron
r(n) = radius of nth orbit
n = Principal quantum number
h = Plank’s Constant
1. Energy
The total energy of an electron remains constant as long as it remains in the same orbit. i.e. it does not radiate energy while revolving around the nucleus.
2. Energy Release
When an electron jumps from a higher orbit having energy ‘En’ to a lower orbit having energy ‘Ep’ then energy is released in the form of energy ‘hv’ i.e.
Eo – Ep = hv = hc / λ
Where,
v = Frequency of Photon
λ = Wavelength of Photon
c = Speed of light
h = Plank’s constant
Qs. Find out the radius, energy and wave number of hydrogen atom with the help of Bohr’s Atomic theory.
HYDROGEN ATOM
A hydrogen atom is the simplest of all atoms. It consist of a proton in the nucleus and an electron revolving around the nucleus.
RADIUS OF HYDROGEN ORBIT
Consider an electron of charge ‘-e’ revolving in a hydrogen atom around a proton of charge ‘+e’ with constant speed v.
When the electron revolves around the nucleus, then two forces balance its motion.
Coulomb’s Force = F = ke² / r² ——– (I)
Centrifugal Force = F = mv² / r ——- (II)
Comparing eq (I) and (II)
ke² / r² = mv² / r
=> ke² / mv² = r² / r
=> r = ke² / mv² ——– (III)
According to Bohr’s theory, angular momentum is an integral multiple of h/2π
mvr = nh / 2π
=> v = nh / 2π mr
=> 1/v = 2π mr / nh
Taking square of both sides
=> 1/v² = 4π² m² r² / n²h²
Substituting the above value in eq (III)
r = Ke² / m = 4π² m² r² / n²h²
=> r / r² = 4π² m k e² / n²h²
=> 1 / r = 4π² m k e² / n²h²
=> r = n²h² / 4π²m k e²
We know that,
k = 1 /4π Єo => 1 / k = 4π Єo
=> r = n²h² / 4 π² m e² x 4π Єo
=> r = n²h² Єo / π e²
The above equation gives the radius of hydrogen atom.
Radii of Various Orbits
Radius of first orbit of hydrogen atom is calculated by substituting the following values in the equation of radius.
n = 1
h = 6.25 x 10(-34) J.sec
m = 9.1 x 10(-31) kg
k = 9 x 10(9) Nm²/col²
e = 1.6 x 10(-19) col
r = (1)² (6.625 x 10(-34)² / 4² (9.1 x 10(-31) (9 x 10(9)) (1.6 x 10-19)²
=> r = 0.53 x 10(-10)m
=> r1 = 0.53 Aº
For other orbits radius is given by
r2 = (2)² x 0.53 Aº
r3 = (3)² x 0.53 Aº
Similarly,
rn = n² x 0.53 Aº
ENERGY OF HYDROGEN ELECTRON
An electron revolving in the orbit of hydrogen atom possesses kinetic energy as well as Potential Energy. Therefore, total energy is given by
E = K.E + P.E —— (I)
Kinetic Energy
When an electron revolves in the orbit, then coulomb’s force is balanced by centrifugal force
ke²/r² = mv²/r
=> mv² = ke²/r
=> 1/2 mv² = ke²/2r
=> K.E = ke²/2r
Potential Energy
Potential energy is given by
P.E = F.dr
=> P.E = Ke² / r² dr
=> P.E = ke² 1 / r² dr
=> P.E = ke² |-1/r|
=> P.E = -ke² [1/r - 1/∞]
=> P.E = -ke² (1/r – 0)
=> P.E = -ke² / r
Total Energy
Substituting the values of K.E and P.E in eq (I)
E = ke² / 2 – ke² / r
=> E = k2² / 2r
Since,
r = n² h²/ 4π² m k e²
=> E = ke² / 2 4π² m k e² / n² h²
=> |E = 2π² m k² e² / n² h²|
The above equation gives the energy of the orbits of hydrogen atom. Negative sign shows that the electron is bound with the nucleus. When energy of the electron becomes positive, then electron will leave the nucleus.
WAVE NUMBER
When art electron jumps from higher orbit to inner orbit, then it radiates energy in the form of photons.
Qs. Explain the spectrum of hydrogen atom.
SPECTRUM OF HYDROGEN ATOM
When an electron jumps from a higher orbit to a lower orbit, it radiates energy which appears in the form of a spectral line. A set of such spectral lines is known as hydrogen spectrum. Hydrogen spectrum is the simplest one which consists of five series.
1. Layman Series
When an electron jumps from a higher orbit to the first orbit, Laymen Series (ultra violet region) is obtained.
The wavelength and wave number of Laymen Series can be calculated by
v = R(H) (1/1² – 1/n²)
Where n = 2, 3, 4, ……
2. Balmer Series
When an electron jumps from a higher orbit to the second orbit then Balmer Series (visible region) is obtained.
The wavelength and wave number of Balmer Series can be calculated by
v = R(H) (1/2² – 1/n²)
Where n = 3, 4, 5, ……
3. Paschen Series
When an electron jumps from a higher orbit to the third orbit then Paschen Series (infra red region) is obtained.
The wavelength and wave number of Paschen Series can be calculated by
v = R(H) (1/3² – 1/n²)
Where n = 4, 5, 6, ……
4. Bracket Series
When an electron jumps from a higher orbit to the fourth orbit then Bracket Series (infra red region) is obtained.
The wavelength and wave number of Bracket Series can be calculated by
v = R(H) (1/4² – 1/n²)
Where n = 5, 6, 7, ……
5. Pfund Series
When an electron jumps from a higher orbit to the fifth orbit then Pfund Series (infra red region) is obtained.
The wavelength and wave number of Pfund Series can be calculated by
v = R(H) (1/5² – 1/n²)
Where n = 6, 7, …..
Qs. Write a note on spectra of X-rays. Also write down the properties.
Introduction
X-Rays were discovered by W.K. Roentgen are also known as Roentgen rays. These rays of shorter wavelength, ranging from 0.1 nm to i nm. X-rays are produced if heavier atoms are bombarded by energetic electrons.
PRODUCTION OF X-RAYS
A Filament F and target T are produced in a vacuum chamber and voltage V is applied across the ends. Electrons are produced by heating the filament. These electrons are accelerated towards the metal by applying very high voltage (several thousands volts). When electrons hit the target, then X-rays are produced. There are two types of spectra obtained from this experiment.
1. A continuous spectrum of frequencies or X-rays Brems Strahlung.
2. Characteristics spectrum or a line spectrum of a limited number of fairly definite frequencies.
1. Continuous Spectra
When electrons hit the metal target, a continuous spectrum of frequencies of X-rays is emitted. The frequencies depend upon the accelerating voltage and are very nearly independent of the material of target.
Continuous spectrum is produced when electrons pass close to the atomic nuclei. The are deflected and slowed down due to which they lose their energy. The energy lost by decelerating electrons appears in the form of photon in the X-ray range. The process is represented as
Atoms + e(Fast) —–> Atom + e(Slow) + hv
2. Characteristic Spectra
In the heavy atoms, electrons are assumed to be arranged in concentric shells at increasing distance from the nucleus. The electrons of inner shell are much tightly bound as compared to the electrons of outer shells. Therefore, a large amount of energy is required to displace them Consequently photons of larger energy are emitted when atoms are stabilized. Thus the transition of inner shell electrons gives rise to high-energy spectra or Characteristic spectra. To obtain characteristic spectra, target metal of higher atomic number is used.
The process of emission of characteristic spectra takes place as follows. When a highly energetic incident electrons knocks an electron from the k-shell, a vacancy occurs in that shell. This vacancy is filled by the arrival of an electron from outside the k-shell, emitting excess amount of energy in the form of photon.
If the electrons jumps only one shell and returns with the emission of X-rays to Y shell, then X-rays are termed as ‘Yα’ X-rays. If the electron jumps two shells and returns with emission of X-rays to suppose ‘Y’ shell, then X-rays are termed as ‘Yβ’ rays and so on, where Y may be K, L, M, ……