Radius of the first Bohr orbit is given by the relation, r1 = 4πε0 (h/2π)²/mee² ------------- Eq-1 Where, Eo = Permittivity of free space h = Planck's constant = 6.63 x 10-34 Js me= Mass of an electron = 9.1 x 10-³¹kg e = Charge of an electron = 1.9 x 10-19 C mp= Mass of a proton = 1.67 x 10 -27 kgRead more
Radius of the first Bohr orbit is given by the relation,
r1 = 4πε0 (h/2π)²/mee² ————- Eq-1
Where,
Eo = Permittivity of free space
h = Planck’s constant = 6.63 x 10-34 Js
me= Mass of an electron = 9.1 x 10-³¹kg
e = Charge of an electron = 1.9 x 10–19 C
mp= Mass of a proton = 1.67 x 10 –27 kg
r= Distance between the electron and the proton
Coulomb attraction between an electron and a proton is given as:
FC = e²/4πε0r²————-Eq-2
Gravitational force of attraction between an electron and a proton is given as:
FG =G mp me /r² ——————- Eq-3 ,
Fc, =-^H- -(3)
Where, G = Gravitational constant = 6.67 x 10⁻¹¹ N m2/kg2
If the electrostatic (Coulomb) force and the gravitational force between an electron and a proton are equal, then we can write:
Therefore , FG=FC
=> G mp me /r² = e²/4πε0r²
Therefore ,
e²/4πε0 =G mp me—————Eq-4
Putting the value of equation (4) in equation (1),we get:
r1= (h/2π)²/G mp m²e
= [6.63 x 10-34/(2 x 3.14)]²/(6.67 x 10⁻¹¹) x (1.67 x 10 –27) x (9.1 x 10-³¹)²
≈1.21 x 10²⁹
It is known that the universe is 156 billion light years wide or 1.5 x 1027 m wide. Hence, we can conclude that the radius of the first Bohr orbit is much greater than the estimated size of the whole universe.
Ans (a). about the same The average angle of deflection of a-particles by a thin gold foil predicted by Thomson’s model is about the same size as predicted by Rutherford's model. This is because the average angle was taken in both models. Ans (b).much less The probability of scattering of a-particleRead more
Ans (a).
about the same
The average angle of deflection of a-particles by a thin gold foil predicted by Thomson’s model is about the same size as predicted by Rutherford’s model. This is because the average angle was taken in both models.
Ans (b).much less
The probability of scattering of a-particles at angles greater than 90° predicted by Thomson’s model is much less than that predicted by Rutherford’s model.
Ans (c).
Scattering is mainly due to single collisions. The chances of a single collision increases linearly with the number of target atoms. Since the number of target atoms increase with an increase in thickness, the collision probability depends linearly on the thickness of the target.
Ans (d).
Thomson’s model
It is wrong to ignore multiple scattering in Thomson’s model for the calculation of average angle of scattering of a-particles by a thin foil. This is because a single collision causes very little deflection in this model. Hence, the observed average scattering angle can be explained only by considering multiple scattering.
Rdius of the orbit of the Earth around the Sun, r= 1.5 x 1011 m Orbital speed of the Earth, v = 3 x 104 m/s Mass of the Earth, m = 6.0 x 10²4 kg According to Bohr's model, angular momentum is quantized and given as: mvr = nh/2π Where, h = Planck's constant = 6.62 x 10⁻34 Js n = Quantum number TherefRead more
Rdius of the orbit of the Earth around the Sun, r= 1.5 x 1011 m
Orbital speed of the Earth, v = 3 x 104 m/s
Mass of the Earth, m = 6.0 x 10²4 kg
According to Bohr’s model, angular momentum is quantized and given as:
mvr = nh/2π
Where,
h = Planck’s constant = 6.62 x 10⁻34 Js
n = Quantum number
Therefore , n = mvr2π/h
= 2π x (6.0 x 10²4 ) x (3 x 104 ) x (1.5 x 1011) /(6.62 x 10⁻34)
= 25.61 x 10⁷³= 2.6 x 10⁷⁴
Hence, the quanta number that characterizes the Earth’ revolution is 2.6 x 1074.
It is given that the energy of the electron beam used to bombard gaseous hydrogen at room temperature is 12.5 eV. Also, the energy of the gaseous hydrogen in its ground state at room temperature is -13.6 eV. When gaseous hydrogen is bombarded with an electron beam, the energy of the gaseous hydrogenRead more
It is given that the energy of the electron beam used to bombard gaseous hydrogen at room temperature is 12.5 eV. Also, the energy of the gaseous hydrogen in its ground state at room temperature is -13.6 eV.
When gaseous hydrogen is bombarded with an electron beam, the energy of the gaseous hydrogen becomes -13.6 + 12.5 eV i.e.,
-1.1 eV.
Orbital energy is related to orbit level (n) as:
E = -13.6/(n)² eV
For n=3 ,
E = -13.6/9 = -1.5 eV
This energy is approximately equal to the energy of gaseous hydrogen. It can be concluded that the electron has jumped from n = 1 to n = 3 level.
During its de-excitation, the electrons can jump from n = 3 to n = 1 directly, which forms a line of the Lyman series of the hydrogen spectrum.
We have the relation for wave number for Lyman series as:
1/λ =Ry (1/1² – 1/n²)
Where,
Ry = Rydberg constant = 1.097 x 107 m⁻1
λ= Wavelength of radiation emitted by the transition of tile electron. For n = 3, we can obtain λ as:
1/λ = 1.097 x 107 (1/1² – 1/3²)
= 1.097 x 107 (1-1/9) = 1.097 x 107 (8/9)
λ =9/8 x ( 1.097 x 107 ) = 102.55 nm
if the electron jumps from n = 2 to n = 1, then the wavelength of the radiation is given as:
1/λ = 1.097 x 107 (1/1² – 1/2²)
= 1.097 x 107 (1-1/4) = 1.097 x 107 (3/4)
λ = 4/( 1.097 x 107 )x(3) = 121.54 nm If the transition takes place from n = 3 to n = 2, then the wavelength of the radiation is given as:
1/λ =1.097 x 107 (1/2²-1/3²)
= 1.097 x 107 (1/4 -1/9) =(1.097 x 107 ) x 5/36
λ= 36/ (5 x 1.097 x 107) = 656.33 nm
This radiation corresponds to the Balmer series of the hydrogen spectrum.
Hence ,in Lyman series ,two wavelengths i.e. 102.5 nm and 121.5nm are emitted .And in the Balmer series, one wavelength i.e.,656.33nm is emitted.
The radius of the innermost orbit of a hydrogen atom, r1 = 5.3 x 10⁻11 m. Let r2 be the radius of the orbit at n = 2. It is related to the radius of the innermost orbit as: r2 = (n)²r1 = 4 x 5.3 x 10⁻¹¹ = 2.12 x 10⁻¹⁰ m For n = 3, we can write the corresponding electron radius as: r3 = (n)²r1 = 9 xRead more
The radius of the innermost orbit of a hydrogen atom, r1 = 5.3 x 10⁻11 m.
Let r2 be the radius of the orbit at n = 2. It is related to the radius of the innermost orbit as:
r2 = (n)²r1
= 4 x 5.3 x 10⁻¹¹ = 2.12 x 10⁻¹⁰ m
For n = 3, we can write the corresponding electron radius as:
r3 = (n)²r1
= 9 x 5.3 x 10⁻¹¹=4.77 x 10⁻10 m
Hence, the radii of an electron for n = 2 and n = 3 orbits are 22.12 x 10⁻¹⁰ m and 4.77 x 10⁻10 m respectively.
The gravitational attraction between electron and proton in a hydrogen atom is weaker than the coulomb attraction by a factor of about 10⁻⁴⁰ .An alternative way of looking at this fact is to estimate the radius of the first Bohr orbit of a hydrogen atom if the electron and proton were bound by gravitational attraction. You will find the answer interesting.
Radius of the first Bohr orbit is given by the relation, r1 = 4πε0 (h/2π)²/mee² ------------- Eq-1 Where, Eo = Permittivity of free space h = Planck's constant = 6.63 x 10-34 Js me= Mass of an electron = 9.1 x 10-³¹kg e = Charge of an electron = 1.9 x 10-19 C mp= Mass of a proton = 1.67 x 10 -27 kgRead more
Radius of the first Bohr orbit is given by the relation,
r1 = 4πε0 (h/2π)²/mee² ————- Eq-1
Where,
Eo = Permittivity of free space
h = Planck’s constant = 6.63 x 10-34 Js
me= Mass of an electron = 9.1 x 10-³¹kg
e = Charge of an electron = 1.9 x 10–19 C
mp= Mass of a proton = 1.67 x 10 –27 kg
r= Distance between the electron and the proton
Coulomb attraction between an electron and a proton is given as:
FC = e²/4πε0r²————-Eq-2
Gravitational force of attraction between an electron and a proton is given as:
FG =G mp me /r² ——————- Eq-3 ,
Fc, =-^H- -(3)
Where, G = Gravitational constant = 6.67 x 10⁻¹¹ N m2/kg2
If the electrostatic (Coulomb) force and the gravitational force between an electron and a proton are equal, then we can write:
Therefore , FG=FC
=> G mp me /r² = e²/4πε0r²
Therefore ,
e²/4πε0 =G mp me—————Eq-4
Putting the value of equation (4) in equation (1),we get:
r1= (h/2π)²/G mp m²e
= [6.63 x 10-34/(2 x 3.14)]²/(6.67 x 10⁻¹¹) x (1.67 x 10 –27) x (9.1 x 10-³¹)²
≈1.21 x 10²⁹
See lessIt is known that the universe is 156 billion light years wide or 1.5 x 1027 m wide. Hence, we can conclude that the radius of the first Bohr orbit is much greater than the estimated size of the whole universe.
Answer the following questions, which help you understand the difference between Thomson’s model and Rutherford’s model better. (a) Is the average angle of deflection of α-particles by a thin gold foil predicted by Thomson’s model much less, about the same, or much greater than that predicted by Rutherford’s model? (b) Is the probability of backward scattering (i.e., scattering of α-particles at angles greater than 90°) predicted by Thomson’s model much less, about the same, or much greater than that predicted by Rutherford’s model? (c) Keeping other factors fixed, it is found experimentally that for small thickness t, the number of α-particles scattered at moderate angles is proportional to t. What clue does this linear dependence on t provide? (d) In which model is it completely wrong to ignore multiple scattering for the calculation of average angle of scattering of α-particles by a thin foil?
Ans (a). about the same The average angle of deflection of a-particles by a thin gold foil predicted by Thomson’s model is about the same size as predicted by Rutherford's model. This is because the average angle was taken in both models. Ans (b).much less The probability of scattering of a-particleRead more
Ans (a).
about the same
The average angle of deflection of a-particles by a thin gold foil predicted by Thomson’s model is about the same size as predicted by Rutherford’s model. This is because the average angle was taken in both models.
Ans (b).much less
The probability of scattering of a-particles at angles greater than 90° predicted by Thomson’s model is much less than that predicted by Rutherford’s model.
Ans (c).
Scattering is mainly due to single collisions. The chances of a single collision increases linearly with the number of target atoms. Since the number of target atoms increase with an increase in thickness, the collision probability depends linearly on the thickness of the target.
Ans (d).
Thomson’s model
It is wrong to ignore multiple scattering in Thomson’s model for the calculation of average angle of scattering of a-particles by a thin foil. This is because a single collision causes very little deflection in this model. Hence, the observed average scattering angle can be explained only by considering multiple scattering.
See lessIn accordance with the Bohr’s model, find the quantum number that characterises the earth’s revolution around the sun in an orbit of radius 1.5 × 10¹¹ m with orbital speed 3 × 10⁴ m/s. (Mass of earth = 6.0 × 10²⁴ kg.)
Rdius of the orbit of the Earth around the Sun, r= 1.5 x 1011 m Orbital speed of the Earth, v = 3 x 104 m/s Mass of the Earth, m = 6.0 x 10²4 kg According to Bohr's model, angular momentum is quantized and given as: mvr = nh/2π Where, h = Planck's constant = 6.62 x 10⁻34 Js n = Quantum number TherefRead more
Rdius of the orbit of the Earth around the Sun, r= 1.5 x 1011 m
Orbital speed of the Earth, v = 3 x 104 m/s
Mass of the Earth, m = 6.0 x 10²4 kg
According to Bohr’s model, angular momentum is quantized and given as:
mvr = nh/2π
Where,
h = Planck’s constant = 6.62 x 10⁻34 Js
n = Quantum number
Therefore , n = mvr2π/h
= 2π x (6.0 x 10²4 ) x (3 x 104 ) x (1.5 x 1011) /(6.62 x 10⁻34)
= 25.61 x 10⁷³= 2.6 x 10⁷⁴
Hence, the quanta number that characterizes the Earth’ revolution is 2.6 x 1074.
See lessA 12.5 eV electron beam is used to bombard gaseous hydrogen at room temperature. What series of wavelengths will be emitted?
It is given that the energy of the electron beam used to bombard gaseous hydrogen at room temperature is 12.5 eV. Also, the energy of the gaseous hydrogen in its ground state at room temperature is -13.6 eV. When gaseous hydrogen is bombarded with an electron beam, the energy of the gaseous hydrogenRead more
It is given that the energy of the electron beam used to bombard gaseous hydrogen at room temperature is 12.5 eV. Also, the energy of the gaseous hydrogen in its ground state at room temperature is -13.6 eV.
When gaseous hydrogen is bombarded with an electron beam, the energy of the gaseous hydrogen becomes -13.6 + 12.5 eV i.e.,
-1.1 eV.
Orbital energy is related to orbit level (n) as:
E = -13.6/(n)² eV
For n=3 ,
E = -13.6/9 = -1.5 eV
This energy is approximately equal to the energy of gaseous hydrogen. It can be concluded that the electron has jumped from n = 1 to n = 3 level.
During its de-excitation, the electrons can jump from n = 3 to n = 1 directly, which forms a line of the Lyman series of the hydrogen spectrum.
We have the relation for wave number for Lyman series as:
1/λ =Ry (1/1² – 1/n²)
Where,
Ry = Rydberg constant = 1.097 x 107 m⁻1
λ= Wavelength of radiation emitted by the transition of tile electron. For n = 3, we can obtain λ as:
1/λ = 1.097 x 107 (1/1² – 1/3²)
= 1.097 x 107 (1-1/9) = 1.097 x 107 (8/9)
λ =9/8 x ( 1.097 x 107 ) = 102.55 nm
if the electron jumps from n = 2 to n = 1, then the wavelength of the radiation is given as:
1/λ = 1.097 x 107 (1/1² – 1/2²)
= 1.097 x 107 (1-1/4) = 1.097 x 107 (3/4)
λ = 4/( 1.097 x 107 )x(3) = 121.54 nm
If the transition takes place from n = 3 to n = 2, then the wavelength of the radiation is given as:
1/λ =1.097 x 107 (1/2²-1/3²)
= 1.097 x 107 (1/4 -1/9) =(1.097 x 107 ) x 5/36
λ= 36/ (5 x 1.097 x 107) = 656.33 nm
This radiation corresponds to the Balmer series of the hydrogen spectrum.
Hence ,in Lyman series ,two wavelengths i.e. 102.5 nm and 121.5nm are emitted .And in the Balmer series, one wavelength i.e.,656.33nm is emitted.
See lessThe radius of the innermost electron orbit of a hydrogen atom is 5.3×10⁻⁻¹¹ m. What are the radii of the n = 2 and n =3 orbits?
The radius of the innermost orbit of a hydrogen atom, r1 = 5.3 x 10⁻11 m. Let r2 be the radius of the orbit at n = 2. It is related to the radius of the innermost orbit as: r2 = (n)²r1 = 4 x 5.3 x 10⁻¹¹ = 2.12 x 10⁻¹⁰ m For n = 3, we can write the corresponding electron radius as: r3 = (n)²r1 = 9 xRead more
The radius of the innermost orbit of a hydrogen atom, r1 = 5.3 x 10⁻11 m.
Let r2 be the radius of the orbit at n = 2. It is related to the radius of the innermost orbit as:
r2 = (n)²r1
= 4 x 5.3 x 10⁻¹¹ = 2.12 x 10⁻¹⁰ m
For n = 3, we can write the corresponding electron radius as:
r3 = (n)²r1
= 9 x 5.3 x 10⁻¹¹=4.77 x 10⁻10 m
Hence, the radii of an electron for n = 2 and n = 3 orbits are 22.12 x 10⁻¹⁰ m and 4.77 x 10⁻10 m respectively.
See less