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.
Suppose 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.
See less