Wavelength of a proton or a neutron, λ ≈ 10-15 m Rest mass energy of an electron: moc2 = 0.511 MeV = 0.511 x 106x 1.6 x 10⁻19 = 0.8176 x 10⁻13 J Planck's constant, h = 6.6 x 10⁻34Js Speed of light, c = 3x 10s m/s The momentum of a proton or a neutron is given as: p=h/λ = (6.6 x 10⁻34)/ 10-15 = 6.6 xRead more
Wavelength of a proton or a neutron, λ ≈ 10-15 m
Rest mass energy of an electron:
moc2 = 0.511 MeV = 0.511 x 106x 1.6 x 10⁻19 = 0.8176 x 10⁻13 J
Planck’s constant, h = 6.6 x 10⁻34Js
Speed of light, c = 3x 10s m/s
The momentum of a proton or a neutron is given as:
p=h/λ = (6.6 x 10⁻34)/ 10-15
= 6.6 x 10⁻19 kg m/s
The relativistic relation for energy (E) is given as:
E = p²c² + m²oc⁴
= (6.6 x 10⁻19 x 3 x 10⁸ )²+ (0.8176 x 10⁻13)²= 392.04 x 10⁻²² + 0.6685 x 10⁻²6
≈392.04 x 10⁻²²
Therefore ,E = 1.98 x 10⁻¹⁰
= ( 1.98 x 10⁻¹⁰ )/(1.6 x 10⁻19) = 1.24 x 10⁹ eV
= 1.24 BeV
Thus, the electron energy emitted from the accelerator at Stanford, USA might be of the order of 1.24 BeV.
Electrons are accelerated by a voltage, V = 50 kV = 50 x 103 V Charge on an electron, e = 1.6 x 10-19 C Mass of an electron, me = 9.11 x 10⁻31 kg Wavelength of yellow light = 5.9 x 10⁻7 m The kinetic energy of the electron is given as: E = eV = 1.6 x 10⁻¹⁹ x 50 x 103 = 8 x 10⁻15 J De Broglie wavelenRead more
Electrons are accelerated by a voltage, V = 50 kV = 50 x 103 V
Charge on an electron, e = 1.6 x 10-19 C
Mass of an electron, me = 9.11 x 10⁻31 kg
Wavelength of yellow light = 5.9 x 10⁻7 m
The kinetic energy of the electron is given as:
E = eV = 1.6 x 10⁻¹⁹ x 50 x 103 = 8 x 10⁻15 J
De Broglie wavelength is given by the relation:
λ = h/√[2meE] = (6.6 x 10⁻34) /√[2 x 9.11 x 10⁻31x 8 x 10⁻15]
= 5.467 x 10⁻¹² m
This wavelength is nearly 105 times less than the wavelength of yellow light.
The resolving power of a microscope is inversely proportional to the wavelength of light used.
Thus, the resolving power of an electron microscope is nearly 105 times that of an optical microscope.
Ans (a). De Broglie wavelength = 2.327 x 10-12 m; neutron is not suitable for the diffraction experiment Kinetic energy of the neutron, K = 150 eV = 150 x 1.6 x 10-19 = 2.4 x 10-17 J Mass of a neutron ,mn=1.675 x 10-27 kg The kinetic energy of the neutron is given by the relation: K = 1/2 mnv2 =>Read more
Ans (a).
De Broglie wavelength = 2.327 x 10-12 m; neutron is not suitable for the diffraction experiment
Kinetic energy of the neutron, K = 150 eV = 150 x 1.6 x 10-19 = 2.4 x 10-17 J
Mass of a neutron ,mn=1.675 x 10-27 kg
The kinetic energy of the neutron is given by the relation:
K = 1/2 mnv2
=> mnv = √(2Kmn)
Where,v = Velocity of the neutron
mnv = Momentum of the neutron
De-Broglie wavelength of the neutron is given as:
λ =h/mnv = h/√(2Kmn)
It is clear that wavelength is inversely proportional to the square root of mass.
Hence. wavelength decreases with increase in mass and vice versa.
λ = (6.6 x 10⁻34)/√ [2(2.4 x 10-17)(1.675 x 10-27)]
= 2.327 x 10⁻¹² m
It is given in the previous problem that the inter-atomic spacing of a crystal is about 1 Aº, i.e., 10-10 m. Hence, the inter-atomic spacing is about a hundred times greater. Hence, a neutron beam of energy 150 eV is not suitable for diffraction experiments.
Ans (b).
De Broglie wavelength = 1.447 x 10-10 m
Room temperature, T = 27°C = 27 + 273 = 300 K
The average kinetic energy of the neutron is given as:
E= 3/2 kT
Where, k = Boltzmann constant = 1.38 x 10⁻23 J mol-1 K⁻1 The wavelength of the neutron is given as:
λ = h/√(2Emn) = h/√(3KmnT )
= (6.6 x 10⁻34)/√[3 x (1.675 x 10⁻27 )x (1.38 x 10⁻23 )x 300]
= 1.447 x 10⁻10 m
This wavelength is comparable to the inter-atomic spacing of a crystal. Hence, the high energy neutron beam should first be thermalised, before using it for diffraction.
An X-ray probe has a greater energy than an electron probe for the same wavelength. Wavelength of light emitted from the probe, λ = 1 Aº = 10⁻¹⁰ m Mass of an electron, me = 9.11 x 10-31 kg Planck's constant, h = 6.6 x 10⁻34 Js Charge on an electron, e = 1.6 x 10-19 C The kinetic energy of the electrRead more
An X-ray probe has a greater energy than an electron probe for the same wavelength.
Wavelength of light emitted from the probe, λ = 1 Aº = 10⁻¹⁰ m
Mass of an electron, me = 9.11 x 10-31 kg
Planck’s constant, h = 6.6 x 10⁻34 Js
Charge on an electron, e = 1.6 x 10-19 C
The kinetic energy of the electron is given as:
E = 1/2 mev²
=>mev = √ (2Eme)
Where,
v = Velocity of the electron
mev = Momentum (p) of the electron
According to the de Broglie principle, the de Broglie wavelength is given as:
λ = h/p = h/mev = h/√ (2Eme)
Therefore , E = h²/(2λ²me) 2
= (6.6 x 10⁻34)² /2 x ( 10⁻¹⁰)² x(9.11 x 10-31 ) = 2.39 x 10⁻¹⁷ J
= ( 2.39 x 10⁻¹⁷ )/(1.6 x 10-19) = 149 .375 eV
Energy of a photon,E’ =hc/λ e eV
= (6.6 x 10⁻34) x (3 x 10⁸ ) /( 10⁻¹⁰) (1.6 x 10-19)
= 12.375 x 10³ eV =12.375 keV
Hence, a photon has a greater energy than an electron for the same wavelength.
Intensity of incident light, I = 10-5 W m⁻2 Surface area of a sodium photocell, A = 2 cm2 = 2 x 10-4 m2 and incident power of the light, P = I x A = 10⁻5 x 2 x 10-4 = 2 x 10-9 W Work function of the metal, φO = 2 eV = 2 x 1.6 x 10-19 = 3.2 x 10-19 J Number of layers of sodium that absorbs the incideRead more
Intensity of incident light, I = 10-5 W m⁻2
Surface area of a sodium photocell, A = 2 cm2 = 2 x 10-4 m2
and incident power of the light, P = I x A = 10⁻5 x 2 x 10-4 = 2 x 10-9 W
Work function of the metal, φO = 2 eV = 2 x 1.6 x 10-19 = 3.2 x 10–19 J
Number of layers of sodium that absorbs the incident energy, n = 5
We know that the effective atomic area of a sodium atom, Ae is 10-20 m2.
Hence, the number of conduction electrons in n layers is given as:
n’ = n x A/Ae = 5 x (2 x 10-4) /(10-20) = 10¹⁷
The incident power is uniformly absorbed by all the electrons continuously. Hence, the amount of energy absorbed per second per electron is:
E = P/n’
= 5 x (2 x 10-9)/(10¹⁷) = 2 x 10⁻²⁶ J/s
Time required for photoelectric emission:
t = φO/E = (3.2 x 10–19) /( 2 x 10⁻²⁶)
=1.6 x 107 s ≈ 0.507 years
The time required for the photoelectric emission is nearly half a year, which is not practical. Hence, the wave picture is in disagreement with the given experiment.
The wavelength of a probe is roughly a measure of the size of a structure that it can probe in some detail. The quark structure of protons and neutrons appears at the minute length-scale of 10⁻¹⁵ m or less. This structure was first probed in early 1970’s using high energy electron beams produced by a linear accelerator at Stanford, USA. Guess what might have been the order of energy of these electron beams. (Rest mass energy of electron = 0.511 MeV.)
Wavelength of a proton or a neutron, λ ≈ 10-15 m Rest mass energy of an electron: moc2 = 0.511 MeV = 0.511 x 106x 1.6 x 10⁻19 = 0.8176 x 10⁻13 J Planck's constant, h = 6.6 x 10⁻34Js Speed of light, c = 3x 10s m/s The momentum of a proton or a neutron is given as: p=h/λ = (6.6 x 10⁻34)/ 10-15 = 6.6 xRead more
Wavelength of a proton or a neutron, λ ≈ 10-15 m
Rest mass energy of an electron:
moc2 = 0.511 MeV = 0.511 x 106x 1.6 x 10⁻19 = 0.8176 x 10⁻13 J
Planck’s constant, h = 6.6 x 10⁻34Js
Speed of light, c = 3x 10s m/s
The momentum of a proton or a neutron is given as:
p=h/λ = (6.6 x 10⁻34)/ 10-15
= 6.6 x 10⁻19 kg m/s
The relativistic relation for energy (E) is given as:
E = p²c² + m²oc⁴
= (6.6 x 10⁻19 x 3 x 10⁸ )²+ (0.8176 x 10⁻13)² = 392.04 x 10⁻²² + 0.6685 x 10⁻²6
≈392.04 x 10⁻²²
Therefore ,E = 1.98 x 10⁻¹⁰
= ( 1.98 x 10⁻¹⁰ )/(1.6 x 10⁻19) = 1.24 x 10⁹ eV
= 1.24 BeV
Thus, the electron energy emitted from the accelerator at Stanford, USA might be of the order of 1.24 BeV.
See lessAn electron microscope uses electrons accelerated by a voltage of 50 kV. Determine the de Broglie wavelength associated with the electrons. If other factors (such as numerical aperture, etc.) are taken to be roughly the same, how does the resolving power of an electron microscope compare with that of an optical microscope which uses yellow light?
Electrons are accelerated by a voltage, V = 50 kV = 50 x 103 V Charge on an electron, e = 1.6 x 10-19 C Mass of an electron, me = 9.11 x 10⁻31 kg Wavelength of yellow light = 5.9 x 10⁻7 m The kinetic energy of the electron is given as: E = eV = 1.6 x 10⁻¹⁹ x 50 x 103 = 8 x 10⁻15 J De Broglie wavelenRead more
Electrons are accelerated by a voltage, V = 50 kV = 50 x 103 V
Charge on an electron, e = 1.6 x 10-19 C
Mass of an electron, me = 9.11 x 10⁻31 kg
Wavelength of yellow light = 5.9 x 10⁻7 m
The kinetic energy of the electron is given as:
E = eV = 1.6 x 10⁻¹⁹ x 50 x 103 = 8 x 10⁻15 J
De Broglie wavelength is given by the relation:
λ = h/√[2meE] = (6.6 x 10⁻34) /√[2 x 9.11 x 10⁻31x 8 x 10⁻15]
= 5.467 x 10⁻¹² m
This wavelength is nearly 105 times less than the wavelength of yellow light.
The resolving power of a microscope is inversely proportional to the wavelength of light used.
Thus, the resolving power of an electron microscope is nearly 105 times that of an optical microscope.
See less(a) Obtain the de Broglie wavelength of a neutron of kinetic energy 150 eV. As you have seen in Exercise 11.31, an electron beam of this energy is suitable for crystal diffraction experiments. Would a neutron beam of the same energy be equally suitable? Explain. (mn = 1.675 × 10⁻²⁷ kg) (b) Obtain the de Broglie wavelength associated with thermal neutrons at room temperature (27 ºC). Hence explain why a fast neutron beam needs to be thermalised with the environment before it can be used for neutron diffraction experiments.
Ans (a). De Broglie wavelength = 2.327 x 10-12 m; neutron is not suitable for the diffraction experiment Kinetic energy of the neutron, K = 150 eV = 150 x 1.6 x 10-19 = 2.4 x 10-17 J Mass of a neutron ,mn=1.675 x 10-27 kg The kinetic energy of the neutron is given by the relation: K = 1/2 mnv2 =>Read more
Ans (a).
De Broglie wavelength = 2.327 x 10-12 m; neutron is not suitable for the diffraction experiment
Kinetic energy of the neutron, K = 150 eV = 150 x 1.6 x 10-19 = 2.4 x 10-17 J
Mass of a neutron ,mn=1.675 x 10-27 kg
The kinetic energy of the neutron is given by the relation:
K = 1/2 mnv2
=> mnv = √(2Kmn)
Where,v = Velocity of the neutron
mnv = Momentum of the neutron
De-Broglie wavelength of the neutron is given as:
λ =h/mnv = h/√(2Kmn)
It is clear that wavelength is inversely proportional to the square root of mass.
Hence. wavelength decreases with increase in mass and vice versa.
λ = (6.6 x 10⁻34)/√ [2(2.4 x 10-17)(1.675 x 10-27)]
= 2.327 x 10⁻¹² m
It is given in the previous problem that the inter-atomic spacing of a crystal is about 1 Aº, i.e., 10-10 m. Hence, the inter-atomic spacing is about a hundred times greater. Hence, a neutron beam of energy 150 eV is not suitable for diffraction experiments.
Ans (b).
De Broglie wavelength = 1.447 x 10-10 m
Room temperature, T = 27°C = 27 + 273 = 300 K
The average kinetic energy of the neutron is given as:
E= 3/2 kT
Where, k = Boltzmann constant = 1.38 x 10⁻23 J mol-1 K⁻1 The wavelength of the neutron is given as:
λ = h/√(2Emn) = h/√(3KmnT )
= (6.6 x 10⁻34)/√[3 x (1.675 x 10⁻27 )x (1.38 x 10⁻23 )x 300]
= 1.447 x 10⁻10 m
This wavelength is comparable to the inter-atomic spacing of a crystal. Hence, the high energy neutron beam should first be thermalised, before using it for diffraction.
See lessCrystal diffraction experiments can be performed using X-rays, or electrons accelerated through appropriate voltage. Which probe has greater energy? (For quantitative comparison, take the wavelength of the probe equal to 1 Å, which is of the order of inter-atomic spacing in the lattice) (me=9.11 × 10⁻³¹ kg).
An X-ray probe has a greater energy than an electron probe for the same wavelength. Wavelength of light emitted from the probe, λ = 1 Aº = 10⁻¹⁰ m Mass of an electron, me = 9.11 x 10-31 kg Planck's constant, h = 6.6 x 10⁻34 Js Charge on an electron, e = 1.6 x 10-19 C The kinetic energy of the electrRead more
An X-ray probe has a greater energy than an electron probe for the same wavelength.
Wavelength of light emitted from the probe, λ = 1 Aº = 10⁻¹⁰ m
Mass of an electron, me = 9.11 x 10-31 kg
Planck’s constant, h = 6.6 x 10⁻34 Js
Charge on an electron, e = 1.6 x 10-19 C
The kinetic energy of the electron is given as:
E = 1/2 mev²
=>mev = √ (2Eme)
Where,
v = Velocity of the electron
mev = Momentum (p) of the electron
According to the de Broglie principle, the de Broglie wavelength is given as:
λ = h/p = h/mev = h/√ (2Eme)
Therefore , E = h²/(2λ²me) 2
= (6.6 x 10⁻34)² /2 x ( 10⁻¹⁰)² x(9.11 x 10-31 ) = 2.39 x 10⁻¹⁷ J
= ( 2.39 x 10⁻¹⁷ )/(1.6 x 10-19) = 149 .375 eV
Energy of a photon,E’ =hc/λ e eV
= (6.6 x 10⁻34) x (3 x 10⁸ ) /( 10⁻¹⁰) (1.6 x 10-19)
= 12.375 x 10³ eV =12.375 keV
Hence, a photon has a greater energy than an electron for the same wavelength.
See lessLight of intensity 10⁻⁵ W m⁻² falls on a sodium photo-cell of surface area 2 cm² . Assuming that the top 5 layers of sodium absorb the incident energy, estimate time required for photoelectric emission in the wave-picture of radiation. The work function for the metal is given to be about 2 eV. What is the implication of your answer?
Intensity of incident light, I = 10-5 W m⁻2 Surface area of a sodium photocell, A = 2 cm2 = 2 x 10-4 m2 and incident power of the light, P = I x A = 10⁻5 x 2 x 10-4 = 2 x 10-9 W Work function of the metal, φO = 2 eV = 2 x 1.6 x 10-19 = 3.2 x 10-19 J Number of layers of sodium that absorbs the incideRead more
Intensity of incident light, I = 10-5 W m⁻2
Surface area of a sodium photocell, A = 2 cm2 = 2 x 10-4 m2
and incident power of the light, P = I x A = 10⁻5 x 2 x 10-4 = 2 x 10-9 W
Work function of the metal, φO = 2 eV = 2 x 1.6 x 10-19 = 3.2 x 10–19 J
Number of layers of sodium that absorbs the incident energy, n = 5
We know that the effective atomic area of a sodium atom, Ae is 10-20 m2.
Hence, the number of conduction electrons in n layers is given as:
n’ = n x A/Ae = 5 x (2 x 10-4) /(10-20) = 10¹⁷
The incident power is uniformly absorbed by all the electrons continuously. Hence, the amount of energy absorbed per second per electron is:
E = P/n’
= 5 x (2 x 10-9)/(10¹⁷) = 2 x 10⁻²⁶ J/s
Time required for photoelectric emission:
t = φO/E = (3.2 x 10–19) /( 2 x 10⁻²⁶)
=1.6 x 107 s ≈ 0.507 years
The time required for the photoelectric emission is nearly half a year, which is not practical. Hence, the wave picture is in disagreement with the given experiment.
See less