Ans (a). Let ν1 be the orbital speed of the electron in a hydrogen atom in the ground state level, n1 = 1. For charge (e) of an electron, ν1 is given by the relation, ν1 = e²/n1 4π ε0 (h/2π) = e²/2 n1ε0h Where, e = 1.6 x 10⁻19 C ε0 = Permittivity of free space = 8.85 x 10⁻12 N-1 C2 m2 h = Planck’s cRead more
Ans (a).
Let ν1 be the orbital speed of the electron in a hydrogen atom in the ground state level, n1 = 1. For charge (e) of an electron, ν1 is given by the relation,
ν1 = e²/n1 4π ε0 (h/2π) = e²/2 n1ε0h
Where, e = 1.6 x 10⁻19 C
ε0 = Permittivity of free space = 8.85 x 10⁻12 N-1 C2 m2
h = Planck’s constant = 6.62 x 10⁻³⁴ Js
Therefore, ν1 = (1.6 x 10⁻19 )2 /2 x (8.85 x 10⁻12) x (6.62 x 10⁻³⁴)
= 0.0218 x 10⁸ = 2.18 x 10⁶ m/s
For level n2= 2 ,we can write the relation for the corresponding orbital speed as:
ν2 = = e²/2 n2ε0h = (1.6 x 10⁻19 )2 /2 x 2 x(8.85 x 10⁻12) x (6.62 x 10⁻³⁴)
=1.09 x 10⁶ m/s
For level n3= 3 ,we can write the relation for the corresponding orbital speed as:
ν3 = = e²/2 n3ε0h = (1.6 x 10⁻19 )2 /2 x 3 x (8.85 x 10⁻12) x (6.62 x 10⁻³⁴)
=7.27 x 10⁵ m/s
Hence, the speed of the electron in a hydrogen atom in n = 1, n=2, and n=3 is 2.18 x 106 m/s, 1.09 x 106 m/s, 7.27 x 105 m/s respectively.
Ans (b).
Let T1 be the orbital period of the electron when it is in level n1 = 1.
Orbital period is related to orbital speed as:
T1 = 2π r/ν1
Where,r1 = Radius of the orbit =( n1)²h²ε0/πme²
h = Planck’s constant = 6.62 x 10⁻³⁴ Js and
e = Charge on an electron = 1.6 x 10⁻19 C.
ε0 = Permittivity of free space = 8.85 x 10⁻12 N⁻1C2 m⁻²
m = Mass of an electron = 9.1 x 10⁻31 kg
Therefore, T1 = 2π r1/ν1
= 2π x (1)² x (6.62 x 10⁻³⁴)² x (8.85 x 10⁻12)/[(2.18 x 10⁶) x π x ( 9.1 x 10⁻31) x (1.6 x 10⁻19)² ]
= 15.27 x 10⁻¹⁷ = 1.527 x 10⁻¹⁶ s
For level n2 = 2, we can write the period as:
T2 = 2π r2/ν2
Where, r2 = Radius of the electron in n2 = 2
= (n2)²h²ε0/πme²
Therefore,T2 =2π r2/ν2
=2π x (2)² x (6.62 x 10⁻³⁴)² x (8.85 x 10⁻12)/[(1.09 x 10⁶) x π x ( 9.1 x 10⁻31) x (1.6 x 10⁻19)² ]
=1.22 x 10⁻¹⁵ s
For level n3 = 3, we can write the period as:
T3 = 2π r3/ν3
Where, r3 = Radius of the electron in n3 = 3
= (n3)²h²ε0/πme²
Therefore,T3 =2π r3/ν3
=2π x (3)² x (6.62 x 10⁻³⁴)² x (8.85 x 10⁻12)/[(7.27 x 10⁵) x π x ( 9.1 x 10⁻31) x (1.6 x 10⁻19)² ]
=4.12 x 10⁻¹⁵ s
Hence, the orbital period in each of these levels is 1.527 x 10⁻¹⁶, 1.22 x 10⁻¹⁵ s, and 4.12 x 10⁻¹⁵ s respectively.
For ground level, n₁ = 1 Let E₁be the energy of this level. It is known that E₁ is related with m as: E₁ =-13.6/n²₁ eV = -13.6 /1² = -13.6 eV The atom is excited to a higher level, n2 = 4. Let E2 be the energy of this level. The amount of energy absorbed by the photon is given as: Therefore E2 =-13.Read more
For ground level, n₁ = 1
Let E₁be the energy of this level. It is known that E₁ is related with m as:
E₁ =-13.6/n²₁ eV
= -13.6 /1² = -13.6 eV
The atom is excited to a higher level, n2 = 4. Let E2 be the energy of this level. The amount of energy absorbed by the photon is given as: Therefore E2 =-13.6/n²2 eV
=-13.6/4² = -13.6 /16 eV
The amount of energy absorbed by the photon is given as :
E = E2-E1
= -13.6 /16 -(-13.6 )
= 13.6 x 15/16 eV
= 13.6 x 15/16 x 1.6 x 10⁻¹⁹ = 2.04 x 10⁻¹⁸ J
For a photon of wavelength λ, the expression of energy is written as:
E = hc/λ
Where,
h = Planck’s constant = 6.6 x 10⁻34 Js and c = Speed of light = 3 x 10⁸ m/s.
Therefore , λ = hc/E = (6.6 x 10⁻34) x (3 x 10⁸ ) /(2.04 x 10⁻¹⁸ )
= 9.7 x 10⁻⁸ m = 97nm
And, frequency of a photon is given by the relation,
ν =c/ λ = (3 x 10⁸)/(9.7 x 10⁻⁸) ≈ 3.1 x 10¹⁵ Hz
Hence, the wavelength of the photon is 97 nm while the frequency is 3.1 x 10¹⁵ Hz.
Ground state energy of hydrogen atom, E = - 13.6 eV This is the total energy of a hydrogen atom. Kinetic energy is equal to the negative of the total energy. Kinetic energy = - E = - (- 13.6) = 13.6 eV Potential energy is equal to the negative of two times of kinetic energy. Potential energy = - 2 xRead more
Ground state energy of hydrogen atom, E = – 13.6 eV
This is the total energy of a hydrogen atom. Kinetic energy is equal to the negative of the total energy.
Kinetic energy = – E = – (- 13.6) = 13.6 eV
Potential energy is equal to the negative of two times of kinetic energy.
Separation of two energy levels in an atom, E = 2.3 eV = 2.3 x 1.6 x 10⁻¹⁹ = 3.68 x 10⁻¹⁹J Let v be the frequency of radiation emitted when the atom transits from the upper level to the lower level. We have the relation for energy as: E = hv Where, h = Plank's constant = 6.62 x 10⁻34 Js Therefore , Read more
Separation of two energy levels in an atom,
E = 2.3 eV = 2.3 x 1.6 x 10⁻¹⁹ = 3.68 x 10⁻¹⁹J
Let v be the frequency of radiation emitted when the atom transits from the upper level to the lower level. We have the relation for energy as: E = hv
Where, h = Plank’s constant = 6.62 x 10⁻34 Js
Therefore , ν= E/h = (3.68 x 10⁻¹⁹)/( 6.62 x 10⁻34) = 5.55 x 1014 Hz Hence, the frequency of the radiation is 5.6 x 1014 Hz.
Rydberg's formula is given as: hc /λ = 21.76 x 10⁻¹⁹ [1/n²₁ -1/n²₂] Where, h = Planck's constant = 6.6 x 10⁻34 Js and c = Speed of light = 3 x 10⁸ m/s. (n₁ and n₂ are integers) The shortest wavelength present in the Paschen series of the spectral lines is given for values n₁=3 and n₂=∝ hc /λ = 21Read more
Rydberg’s formula is given as:
hc /λ = 21.76 x 10⁻¹⁹ [1/n²₁ -1/n²₂]
Where,
h = Planck’s constant = 6.6 x 10⁻34 Js and c = Speed of light = 3 x 10⁸ m/s. (n₁ and n₂ are integers)
The shortest wavelength present in the Paschen series of the spectral lines is given for values n₁=3 and n₂=∝
hc /λ = 21.76 x 10⁻¹⁹ [1/(3)² -1/(∝) ²]
λ = (6.6 x 10⁻34 ) x (3 x 10⁸) x 9 ]/( 21.76 x 10⁻¹⁹)
In the alpha-particle scattering experiment, if a thin sheet of solid hydrogen is used in place of a gold foil, then the scattering angle would not be large enough. This is because the mass of hydrogen (1.67 x 1 0⁻27 kg) is less than the mass of incident α-particles (6.64 x 10⁻27 kg). Thus, the massRead more
In the alpha-particle scattering experiment, if a thin sheet of solid hydrogen is used in place of a gold foil, then the scattering angle would not be large enough. This is because the mass of hydrogen (1.67 x 1 0⁻27 kg) is less than the mass of incident α-particles (6.64 x 10⁻27 kg). Thus, the mass of the scattering particle is more than the target nucleus (hydrogen). As a result, the α-particles would not bounce back if solid hydrogen is used in the a-particle scattering experiment
Ans (a). No different from The sizes of the atoms taken in Thomson's model and Rutherford’s model have the same order of magnitude. Ans (b). In the ground state of Thomson’s model, the electrons are in stable equilibrium. However, in Rutherford's model, the electrons always experience a net force. ARead more
Ans (a).
No different from
The sizes of the atoms taken in Thomson’s model and Rutherford’s model have the same order of magnitude.
Ans (b).
In the ground state of Thomson’s model, the electrons are in stable equilibrium.
However, in Rutherford’s model, the electrons always experience a net force.
Ans (c).
A classical atom based on Rutherford’s model is doomed to collapse.
Ans (d).
An atom has a nearly continuous mass distribution in Thomson’s model, but has a highly non-uniform mass distribution in Rutherford’s model.
Ans (e).
The positively charged part of the atom possesses most of the mass in both the models.
Ans (a). Quarks inside protons and neutrons carry fractional charges. This is because nuclear force increases extremely if they are pulled apart. Therefore, fractional charges may exist in nature; observable charges are still the integral multiple of an electrical charge. Ans (b). The basic relationRead more
Ans (a).
Quarks inside protons and neutrons carry fractional charges. This is because nuclear force increases extremely if they are pulled apart. Therefore, fractional charges may exist in nature; observable charges are still the integral multiple of an electrical charge.
Ans (b).
The basic relations for electric field and magnetic field are
(eV = 1/2 mv²) and ( eBv = mv²/r) respectively
These relations include e (electric charge) , v (velocity), m (mass ) ,V (potential ),r (radius ), and B (magnetic field). These relations give the value of velocity of an electron as
{v = √[2V (e/m)]} and { v = Br(e/m)} respectively.
It can be observed from these relations that the dynamics of an electron is determined not by e and m separately ,but by the ration e/m.
Ans (c).
At atmospheric pressure, the ions of gases have no chance of reaching their respective electrons because of collision and recombination with other gas molecules. Hence, gases are insulators at atmospheric pressure. At low pressures, ions have a chance of reaching their respective electrodes and constitute a current. Hence, they conduct electricity at these pressures.
Ans (d).
The work function of a metal is the minimum energy required for a conduction electron to get out of the metal surface. All the electrons in an atom do not have the same energy level. When a ray having some photon energy is incident on a metal surface, the electrons come out from different levels with different energies. Hence, these emitted electrons show different energy distributions.
Ans (e).The absolute value of energy of a particle is arbitrary within the additive constant. Hence, wavelength (λ) is significant, but the frequency (v) associated with an electron has no direct physical significance.
Therefore, the product vλ (phase speed) has no physical significance.
Temperature, T = 27C = 27º+ 273 =300K Mean separation between two electrons , r = 2 x 10⁻¹⁰ m De Broglie wavelength of an electron is given as : λ = h/ √ [3mkT] Where ,h = Planck’s constant = 6.6 x 10-34 Js m = Mass of an electron = 9.11 x 10-31 kg k= Boltzmann constant = 1.38 x 10⁻23 J mol-1 K-1 ThRead more
Temperature, T = 27C = 27º+ 273 =300K
Mean separation between two electrons , r = 2 x 10⁻¹⁰ m
De Broglie wavelength of an electron is given as :
λ = h/ √ [3mkT]
Where ,h = Planck’s constant = 6.6 x 10-34 Js
m = Mass of an electron = 9.11 x 10-31 kg
k= Boltzmann constant = 1.38 x 10⁻23 J mol-1 K-1
Therefore ,λ = (6.6 x 10-34)/ √ [3 x (9.11 x 10-31) x (1.38 x 10⁻23) x 300]
≈ 6.2 x 10 ⁻⁹ m
Hence, the de Broglie wavelength is much greater than the given inter-electron separation.
De Broglie wavelength associated with He atom = 0.7268 x 10-10 m Room temperature, T = 27°C = 27 + 273 = 300 K Atmospheric pressure, P = 1 atm = 1.01 x 10⁵ Pa Atomic weight of a He atom = 4 Avogadro's number, Na = 6.023 x 1023 Boltzmann constant, k= 1.38 x 10-23 J mol-1 K-1 Average energy of a gas aRead more
De Broglie wavelength associated with He atom = 0.7268 x 10-10 m
Room temperature, T = 27°C = 27 + 273 = 300 K
Atmospheric pressure, P = 1 atm = 1.01 x 10⁵ Pa
Atomic weight of a He atom = 4
Avogadro’s number, Na = 6.023 x 1023
Boltzmann constant, k= 1.38 x 10-23 J mol-1 K-1
Average energy of a gas at temperature T, is given as:
E= 3/2 kT
De Broglie wavelength is given by the relation:
λ = h/√[2mE]
Where,
m = Mass of a He atom
= Atomic weight /Na = 4/(6.023 x 1023)
=6.64 x 10⁻²⁴ g = 6.64 x 10⁻²⁷
Therefore , λ = h/√[3mkT]
= (6.6 x 10⁻34) /√[3 x (6.64 x 10⁻²⁷) x (1.38 x 10-23) x 300]
= 0.7268 x 10⁻¹⁰ m
We have the ideal gas formula:
PV = RT
PV = kNT
V/N = kT/P_
Where,
V = Volume of the gas
N = Number of moles of the gas
Mean separation between two atoms of the gas is given by the relation:
r³ = (V/N) = (kT/P) = [(1.38 x 10-23) x (300) /( 1.01 x 10⁵)]
r = 3.35 x 10⁻⁹ m
Hence, the mean separation between the atoms is much greater than the de Broglie wavelength.
(a) Using the Bohr’s model calculate the speed of the electron in a hydrogen atom in the n = 1, 2, and 3 levels. (b) Calculate the orbital period in each of these levels.
Ans (a). Let ν1 be the orbital speed of the electron in a hydrogen atom in the ground state level, n1 = 1. For charge (e) of an electron, ν1 is given by the relation, ν1 = e²/n1 4π ε0 (h/2π) = e²/2 n1ε0h Where, e = 1.6 x 10⁻19 C ε0 = Permittivity of free space = 8.85 x 10⁻12 N-1 C2 m2 h = Planck’s cRead more
Ans (a).
Let ν1 be the orbital speed of the electron in a hydrogen atom in the ground state level, n1 = 1. For charge (e) of an electron, ν1 is given by the relation,
ν1 = e²/n1 4π ε0 (h/2π) = e²/2 n1ε0h
Where, e = 1.6 x 10⁻19 C
ε0 = Permittivity of free space = 8.85 x 10⁻12 N-1 C2 m2
h = Planck’s constant = 6.62 x 10⁻³⁴ Js
Therefore, ν1 = (1.6 x 10⁻19 )2 /2 x (8.85 x 10⁻12) x (6.62 x 10⁻³⁴)
= 0.0218 x 10⁸ = 2.18 x 10⁶ m/s
For level n2= 2 ,we can write the relation for the corresponding orbital speed as:
ν2 = = e²/2 n2ε0h = (1.6 x 10⁻19 )2 /2 x 2 x(8.85 x 10⁻12) x (6.62 x 10⁻³⁴)
=1.09 x 10⁶ m/s
For level n3= 3 ,we can write the relation for the corresponding orbital speed as:
ν3 = = e²/2 n3ε0h = (1.6 x 10⁻19 )2 /2 x 3 x (8.85 x 10⁻12) x (6.62 x 10⁻³⁴)
=7.27 x 10⁵ m/s
Hence, the speed of the electron in a hydrogen atom in n = 1, n=2, and n=3 is 2.18 x 106 m/s, 1.09 x 106 m/s, 7.27 x 105 m/s respectively.
Ans (b).
Let T1 be the orbital period of the electron when it is in level n1 = 1.
Orbital period is related to orbital speed as:
T1 = 2π r/ν1
Where,r1 = Radius of the orbit =( n1)²h²ε0/πme²
h = Planck’s constant = 6.62 x 10⁻³⁴ Js and
e = Charge on an electron = 1.6 x 10⁻19 C.
ε0 = Permittivity of free space = 8.85 x 10⁻12 N⁻1C2 m⁻²
m = Mass of an electron = 9.1 x 10⁻31 kg
Therefore, T1 = 2π r1/ν1
= 2π x (1)² x (6.62 x 10⁻³⁴)² x (8.85 x 10⁻12)/[(2.18 x 10⁶) x π x ( 9.1 x 10⁻31) x (1.6 x 10⁻19)² ]
= 15.27 x 10⁻¹⁷ = 1.527 x 10⁻¹⁶ s
For level n2 = 2, we can write the period as:
T2 = 2π r2/ν2
Where, r2 = Radius of the electron in n2 = 2
= (n2)²h²ε0/πme²
Therefore,T2 =2π r2/ν2
=2π x (2)² x (6.62 x 10⁻³⁴)² x (8.85 x 10⁻12)/[(1.09 x 10⁶) x π x ( 9.1 x 10⁻31) x (1.6 x 10⁻19)² ]
=1.22 x 10⁻¹⁵ s
For level n3 = 3, we can write the period as:
T3 = 2π r3/ν3
Where, r3 = Radius of the electron in n3 = 3
= (n3)²h²ε0/πme²
Therefore,T3 =2π r3/ν3
=2π x (3)² x (6.62 x 10⁻³⁴)² x (8.85 x 10⁻12)/[(7.27 x 10⁵) x π x ( 9.1 x 10⁻31) x (1.6 x 10⁻19)² ]
=4.12 x 10⁻¹⁵ s
Hence, the orbital period in each of these levels is 1.527 x 10⁻¹⁶, 1.22 x 10⁻¹⁵ s, and 4.12 x 10⁻¹⁵ s respectively.
See lessA hydrogen atom initially in the ground level absorbs a photon, which excites it to the n = 4 level. Determine the wavelength and frequency of photon.
For ground level, n₁ = 1 Let E₁be the energy of this level. It is known that E₁ is related with m as: E₁ =-13.6/n²₁ eV = -13.6 /1² = -13.6 eV The atom is excited to a higher level, n2 = 4. Let E2 be the energy of this level. The amount of energy absorbed by the photon is given as: Therefore E2 =-13.Read more
For ground level, n₁ = 1
Let E₁be the energy of this level. It is known that E₁ is related with m as:
E₁ =-13.6/n²₁ eV
= -13.6 /1² = -13.6 eV
The atom is excited to a higher level, n2 = 4. Let E2 be the energy of this level.
The amount of energy absorbed by the photon is given as:
Therefore E2 =-13.6/n²2 eV
=-13.6/4² = -13.6 /16 eV
The amount of energy absorbed by the photon is given as :
E = E2-E1
= -13.6 /16 -(-13.6 )
= 13.6 x 15/16 eV
= 13.6 x 15/16 x 1.6 x 10⁻¹⁹ = 2.04 x 10⁻¹⁸ J
For a photon of wavelength λ, the expression of energy is written as:
E = hc/λ
Where,
h = Planck’s constant = 6.6 x 10⁻34 Js and c = Speed of light = 3 x 10⁸ m/s.
Therefore , λ = hc/E = (6.6 x 10⁻34) x (3 x 10⁸ ) /(2.04 x 10⁻¹⁸ )
= 9.7 x 10⁻⁸ m = 97nm
And, frequency of a photon is given by the relation,
ν =c/ λ = (3 x 10⁸)/(9.7 x 10⁻⁸) ≈ 3.1 x 10¹⁵ Hz
Hence, the wavelength of the photon is 97 nm while the frequency is 3.1 x 10¹⁵ Hz.
See lessThe ground state energy of hydrogen atom is –13.6 eV. What are the kinetic and potential energies of the electron in this state?
Ground state energy of hydrogen atom, E = - 13.6 eV This is the total energy of a hydrogen atom. Kinetic energy is equal to the negative of the total energy. Kinetic energy = - E = - (- 13.6) = 13.6 eV Potential energy is equal to the negative of two times of kinetic energy. Potential energy = - 2 xRead more
Ground state energy of hydrogen atom, E = – 13.6 eV
This is the total energy of a hydrogen atom. Kinetic energy is equal to the negative of the total energy.
Kinetic energy = – E = – (- 13.6) = 13.6 eV
Potential energy is equal to the negative of two times of kinetic energy.
Potential energy = – 2 x (13.6) = -27.2 eV
See lessA difference of 2.3 eV separates two energy levels in an atom. What is the frequency of radiation emitted when the atom make a transition from the upper level to the lower level?
Separation of two energy levels in an atom, E = 2.3 eV = 2.3 x 1.6 x 10⁻¹⁹ = 3.68 x 10⁻¹⁹J Let v be the frequency of radiation emitted when the atom transits from the upper level to the lower level. We have the relation for energy as: E = hv Where, h = Plank's constant = 6.62 x 10⁻34 Js Therefore , Read more
Separation of two energy levels in an atom,
E = 2.3 eV = 2.3 x 1.6 x 10⁻¹⁹ = 3.68 x 10⁻¹⁹J
Let v be the frequency of radiation emitted when the atom transits from the upper level to the lower level. We have the relation for energy as: E = hv
Where, h = Plank’s constant = 6.62 x 10⁻34 Js
Therefore , ν= E/h = (3.68 x 10⁻¹⁹)/( 6.62 x 10⁻34) = 5.55 x 1014 Hz
See lessHence, the frequency of the radiation is 5.6 x 1014 Hz.
What is the shortest wavelength present in the Paschen series of spectral lines?
Rydberg's formula is given as: hc /λ = 21.76 x 10⁻¹⁹ [1/n²₁ -1/n²₂] Where, h = Planck's constant = 6.6 x 10⁻34 Js and c = Speed of light = 3 x 10⁸ m/s. (n₁ and n₂ are integers) The shortest wavelength present in the Paschen series of the spectral lines is given for values n₁=3 and n₂=∝ hc /λ = 21Read more
Rydberg’s formula is given as:
hc /λ = 21.76 x 10⁻¹⁹ [1/n²₁ -1/n²₂]
Where,
h = Planck’s constant = 6.6 x 10⁻34 Js and c = Speed of light = 3 x 10⁸ m/s. (n₁ and n₂ are integers)
The shortest wavelength present in the Paschen series of the spectral lines is given for values n₁=3 and n₂=∝
hc /λ = 21.76 x 10⁻¹⁹ [1/(3)² -1/(∝) ²]
λ = (6.6 x 10⁻34 ) x (3 x 10⁸) x 9 ]/( 21.76 x 10⁻¹⁹)
= 8.189 x 10⁻⁷m = 818.9 nn
See lessSuppose you are given a chance to repeat the alpha-particle scattering experiment using a thin sheet of solid hydrogen in place of the gold foil. (Hydrogen is a solid at temperatures below 14 K.) What results do you expect?
In the alpha-particle scattering experiment, if a thin sheet of solid hydrogen is used in place of a gold foil, then the scattering angle would not be large enough. This is because the mass of hydrogen (1.67 x 1 0⁻27 kg) is less than the mass of incident α-particles (6.64 x 10⁻27 kg). Thus, the massRead more
In the alpha-particle scattering experiment, if a thin sheet of solid hydrogen is used in place of a gold foil, then the scattering angle would not be large enough. This is because the mass of hydrogen (1.67 x 1 0⁻27 kg) is less than the mass of incident α-particles (6.64 x 10⁻27 kg). Thus, the mass of the scattering particle is more than the target nucleus (hydrogen). As a result, the α-particles would not bounce back if solid hydrogen is used in the a-particle scattering experiment
See lessChoose the correct alternative from the clues given at the end of the each statement: (a) The size of the atom in Thomson’s model is ………. the atomic size in Rutherford’s model. (much greater than/no different from/much less than.) (b) In the ground state of ………. electrons are in stable equilibrium, while in ………. electrons always experience a net force. (Thomson’s model/ Rutherford’s model.) (c) A classical atom based on ………. is doomed to collapse. (Thomson’s model/ Rutherford’s model.) (d) An atom has a nearly continuous mass distribution in a ………. but has a highly non-uniform mass distribution in ………. (Thomson’s model/ Rutherford’s model.) (e) The positively charged part of the atom possesses most of the mass in ………. (Rutherford’s model/both the models.)
Ans (a). No different from The sizes of the atoms taken in Thomson's model and Rutherford’s model have the same order of magnitude. Ans (b). In the ground state of Thomson’s model, the electrons are in stable equilibrium. However, in Rutherford's model, the electrons always experience a net force. ARead more
Ans (a).
No different from
The sizes of the atoms taken in Thomson’s model and Rutherford’s model have the same order of magnitude.
Ans (b).
In the ground state of Thomson’s model, the electrons are in stable equilibrium.
However, in Rutherford’s model, the electrons always experience a net force.
Ans (c).
A classical atom based on Rutherford’s model is doomed to collapse.
Ans (d).
An atom has a nearly continuous mass distribution in Thomson’s model, but has a highly non-uniform mass distribution in Rutherford’s model.
Ans (e).
The positively charged part of the atom possesses most of the mass in both the models.
See lessAnswer the following questions: (a) Quarks inside protons and neutrons are thought to carry fractional charges [(+2/3)e ; (–1/3)e]. Why do they not show up in Millikan’s oil-drop experiment? (b) What is so special about the combination e/m? Why do we not simply talk of e and m separately? (c) Why should gases be insulators at ordinary pressures and start conducting at very low pressures? (d) Every metal has a definite work function. Why do all photoelectrons not come out with the same energy if incident radiation is monochromatic? Why is there an energy distribution of photoelectrons? (e) The energy and momentum of an electron are related to the frequency and wavelength of the associated matter wave by the relations: E = h ν, p = h/λ But while the value of λ is physically significant, the value of ν (and therefore, the value of the phase speed ν λ) has no physical significance. Why?
Ans (a). Quarks inside protons and neutrons carry fractional charges. This is because nuclear force increases extremely if they are pulled apart. Therefore, fractional charges may exist in nature; observable charges are still the integral multiple of an electrical charge. Ans (b). The basic relationRead more
Ans (a).
Quarks inside protons and neutrons carry fractional charges. This is because nuclear force increases extremely if they are pulled apart. Therefore, fractional charges may exist in nature; observable charges are still the integral multiple of an electrical charge.
Ans (b).
The basic relations for electric field and magnetic field are
(eV = 1/2 mv²) and ( eBv = mv²/r) respectively
These relations include e (electric charge) , v (velocity), m (mass ) ,V (potential ),r (radius ), and B (magnetic field). These relations give the value of velocity of an electron as
{v = √[2V (e/m)]} and { v = Br(e/m)} respectively.
It can be observed from these relations that the dynamics of an electron is determined not by e and m separately ,but by the ration e/m.
Ans (c).
At atmospheric pressure, the ions of gases have no chance of reaching their respective electrons because of collision and recombination with other gas molecules. Hence, gases are insulators at atmospheric pressure. At low pressures, ions have a chance of reaching their respective electrodes and constitute a current. Hence, they conduct electricity at these pressures.
Ans (d).
The work function of a metal is the minimum energy required for a conduction electron to get out of the metal surface. All the electrons in an atom do not have the same energy level. When a ray having some photon energy is incident on a metal surface, the electrons come out from different levels with different energies. Hence, these emitted electrons show different energy distributions.
Ans (e).The absolute value of energy of a particle is arbitrary within the additive constant. Hence, wavelength (λ) is significant, but the frequency (v) associated with an electron has no direct physical significance.
Therefore, the product vλ (phase speed) has no physical significance.
Group speed is given as:
vG= vdv/dk = dv/d (1/λ) = dE/dp = d (p²/2m) /dp = p/m
dv dE
This quantity has a physical meaning.
See lessCompute the typical de Broglie wavelength of an electron in a metal at 27 ºC and compare it with the mean separation between two electrons in a metal which is given to be about 2 × 10⁻¹⁰ m. [Note: Exercises 11.35 and 11.36 reveal that while the wave-packets associated with gaseous molecules under ordinary conditions are non-overlapping, the electron wave-packets in a metal strongly overlap with one another. This suggests that whereas molecules in an ordinary gas can be distinguished apart, electrons in a metal cannot be distintguished apart from one another. This indistinguishibility has many fundamental implications which you will explore in more advanced Physics courses.]
Temperature, T = 27C = 27º+ 273 =300K Mean separation between two electrons , r = 2 x 10⁻¹⁰ m De Broglie wavelength of an electron is given as : λ = h/ √ [3mkT] Where ,h = Planck’s constant = 6.6 x 10-34 Js m = Mass of an electron = 9.11 x 10-31 kg k= Boltzmann constant = 1.38 x 10⁻23 J mol-1 K-1 ThRead more
Temperature, T = 27C = 27º+ 273 =300K
Mean separation between two electrons , r = 2 x 10⁻¹⁰ m
De Broglie wavelength of an electron is given as :
λ = h/ √ [3mkT]
Where ,h = Planck’s constant = 6.6 x 10-34 Js
m = Mass of an electron = 9.11 x 10-31 kg
k= Boltzmann constant = 1.38 x 10⁻23 J mol-1 K-1
Therefore ,λ = (6.6 x 10-34)/ √ [3 x (9.11 x 10-31) x (1.38 x 10⁻23) x 300]
≈ 6.2 x 10 ⁻⁹ m
Hence, the de Broglie wavelength is much greater than the given inter-electron separation.
See lessFind the typical de Broglie wavelength associated with a He atom in helium gas at room temperature (27 ºC) and 1 atm pressure; and compare it with the mean separation between two atoms under these conditions.
De Broglie wavelength associated with He atom = 0.7268 x 10-10 m Room temperature, T = 27°C = 27 + 273 = 300 K Atmospheric pressure, P = 1 atm = 1.01 x 10⁵ Pa Atomic weight of a He atom = 4 Avogadro's number, Na = 6.023 x 1023 Boltzmann constant, k= 1.38 x 10-23 J mol-1 K-1 Average energy of a gas aRead more
De Broglie wavelength associated with He atom = 0.7268 x 10-10 m
Room temperature, T = 27°C = 27 + 273 = 300 K
Atmospheric pressure, P = 1 atm = 1.01 x 10⁵ Pa
Atomic weight of a He atom = 4
Avogadro’s number, Na = 6.023 x 1023
Boltzmann constant, k= 1.38 x 10-23 J mol-1 K-1
Average energy of a gas at temperature T, is given as:
E= 3/2 kT
De Broglie wavelength is given by the relation:
λ = h/√[2mE]
Where,
m = Mass of a He atom
= Atomic weight /Na = 4/(6.023 x 1023)
=6.64 x 10⁻²⁴ g = 6.64 x 10⁻²⁷
Therefore , λ = h/√[3mkT]
= (6.6 x 10⁻34) /√[3 x (6.64 x 10⁻²⁷) x (1.38 x 10-23) x 300]
= 0.7268 x 10⁻¹⁰ m
We have the ideal gas formula:
PV = RT
PV = kNT
V/N = kT/P_
Where,
V = Volume of the gas
N = Number of moles of the gas
Mean separation between two atoms of the gas is given by the relation:
r³ = (V/N) = (kT/P) = [(1.38 x 10-23) x (300) /( 1.01 x 10⁵)]
r = 3.35 x 10⁻⁹ m
Hence, the mean separation between the atoms is much greater than the de Broglie wavelength.
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