Ans (a). Wavelength produced by an X-ray tube, λ = 0.45 Aº = 0.45 x 10-10 m Planck’s constant, h = 6.626 x 10⁻34Js Speed of light, c = 3 x 108 m/s The maximum energy of a photon is given as: E = hc/λ = (6.626 x 10⁻34) ( 3 x 108 ) /(0.45 x 10-10) (1.6 x 10¹⁹ ) = 27.6x10³ eV = 27.6 keV Therefore, theRead more
Ans (a).
Wavelength produced by an X-ray tube, λ = 0.45 Aº = 0.45 x 10-10 m
Planck’s constant, h = 6.626 x 10⁻34Js
Speed of light, c = 3 x 108 m/s
The maximum energy of a photon is given as:
E = hc/λ = (6.626 x 10⁻34) ( 3 x 108 ) /(0.45 x 10-10) (1.6 x 10¹⁹ )
= 27.6×10³ eV = 27.6 keV
Therefore, the maximum energy of an X-ray photon is 27.6 keV.
Ans (b).
Accelerating voltage provides energy to the electrons for producing X-rays. To get an X-ray of 27.6 keV, the incident electrons must possess at least 27.6 keV of kinetic electric energy. Hence, an accelerating voltage of the order of 30 keV is required for producing X-rays.
Potential of an anode, V = 100 V Magnetic field experienced by the electrons, B = 2.83 x 10-4 T Radius of the circular orbit r = 12.0 cm = 12.0 x 10-2 m Mass of each electron = m and charge on each electron = e Velocity of each electron = v The energy of each electron is equal to its kinetic energy,Read more
Potential of an anode, V = 100 V
Magnetic field experienced by the electrons, B = 2.83 x 10-4 T
Radius of the circular orbit r = 12.0 cm = 12.0 x 10-2 m
Mass of each electron = m and
charge on each electron = e
Velocity of each electron = v
The energy of each electron is equal to its kinetic energy, i.e.,
1/2 mv² = eV
v²= 2eV/m————- Eq-1
It is the magnetic field, due to its bending nature, that provides the centripetal force (F = mv²/r) Centripetal force = Magnetic force
mv²/r = evB
eB = mv/r
v = eBr/m ——Eq-2
Putting the value of v in equation (1), we get:
2eV/m = e²B²r²/m²
e/m = 2V/B²r²
= 2 x 100 / (2.83 x 10-4 ) x (12.0 x 10-2) = 1.73 x 10¹¹Ckg-1
Therefore, the specific charge ratio (e/m) is 1.73 x 10¹¹ Ckg-1.
Ans (a). Speed of an electron, v = 5.20 x 106 m/s Magnetic field experienced by the electron, B = 1.30 x 10-4 T Specific charge of an electron, e/m = 1.76 x 1011 C kg-1 Where, e = Charge on the electron = 1.6 x 10-19 C m = Mass of the electron = 9.1 x 10-31 kg-1 The force exerted on the electron isRead more
Ans (a).
Speed of an electron, v = 5.20 x 106 m/s
Magnetic field experienced by the electron, B = 1.30 x 10–4 T
Specific charge of an electron, e/m = 1.76 x 1011 C kg-1
Where, e = Charge on the electron = 1.6 x 10-19 C
m = Mass of the electron = 9.1 x 10–31 kg-1
The force exerted on the electron is given as:
F = e|v x B|
= evB sin 0
0 = Angle between the magnetic field and the beam velocity.
The magnetic field is normal to the direction of beam.
Therefore ,0 = 90°
F=evB …(1)
The beam traces a circular path of radius, r. It is the magnetic field, due to its bending nature, that provides the centripetal force (F = mv²/r) for the beam.
Hence, equation (1) reduces to:
evB = mv²/r
Therefore , r = mv/eB = v/(e/m)B
= (5.20 x 106)/(1.76 x 1011) x ( 1.30 x 10–4 ) =0.227 m =22.7 cm
Therefore, the radius of the circular path is 22.7 cm.
Ans (b).
Energy of the electron beam, E = 20 MeV = 20 x 106 x 1.6 x 10⁻19 J
The energy of the electron is given as:
E = 1/2 mv²
Therefore , v = √(2E/m)
= √ (2 x 20 x 106 x 1.6 x 10⁻19 )/(9.1 x 10–31) =2.652 x 10⁹ m/s
This result is incorrect because nothing can move faster than light. In the above formula, the expression (mv2/2) for energy can only be used in the non-relativistic limit, i.e., for v « c.
When very high speeds are concerned, the relativistic domain comes into consideration.
In the relativistic domain, mass is given as:
m = m0 √[1 – v²/c² ]
Where,
m0 = Mass of the particle at rest
Hence, the radius of the circular path is given as:
Ans (a). Potential difference across the evacuated tube, V = 500 V Specific charge of an electron, e/m = 1.76 x 1011 C kg-1 The speed of each emitted electron is given by the relation for kinetic energy as: KE = 1/2 mv²=eV Therefore ,v =√(2eV/m) = √(2V e/m) = √(2x 500 x 1.76 x 10¹¹) =1.327 x 107 m/sRead more
Ans (a).
Potential difference across the evacuated tube, V = 500 V
Specific charge of an electron, e/m = 1.76 x 1011 C kg-1
The speed of each emitted electron is given by the relation for kinetic energy as:
KE = 1/2 mv²=eV
Therefore ,v =√(2eV/m) = √(2V e/m)
= √(2x 500 x 1.76 x 10¹¹) =1.327 x 107 m/s
Therefore, the speed of each emitted electron is 1.327 x 107 m/s.
Ans (b).
Potential of the anode, V = 10 MV = 10 x 106 V
The speed of each electron is given as:
v = √ (2V e/m))
= √ (2 x 107 x 1.76x 10¹¹)- = 1.88 x 109 m/s
This result is wrong because nothing can move faster than light. In the above formula, the expression (mv2/2) for energy can only be used in the non-relativistic limit, i.e., for v « c.
For very high speed problems, relativistic equations must be considered for solving them. In the relativistic limit, the total energy is given as: E = me2
Where, m = Relativistic mass = mo √ (1 – v2/c2)
mo = Mass of the particle at rest Kinetic energy is given as: K = mc2 – moc2
Temperature of the nitrogen molecule, T = 300 K Atomic mass of nitrogen = 14.0076 u Hence, mass of the nitrogen molecule, m = 2 x 14.0076 = 28.0152 u But 1 u = 1.66 x 10⁻27 kg Therefore, m = 28.0152 x 1.66 x 10⁻27 kg Planck’s constant, h = 6.63 x 10⁻34 Js Boltzmann constant, k = 1.38 x 10⁻23J K⁻1 WeRead more
Temperature of the nitrogen molecule, T = 300 K
Atomic mass of nitrogen = 14.0076 u
Hence, mass of the nitrogen molecule, m = 2 x 14.0076 = 28.0152 u
But 1 u = 1.66 x 10⁻27 kg
Therefore, m = 28.0152 x 1.66 x 10⁻27 kg
Planck’s constant, h = 6.63 x 10⁻34 Js
Boltzmann constant, k = 1.38 x 10⁻23J K⁻1
We have the expression that relates mean kinetic energy (3KT/2 )of the nitrogen molecule with the root mean square speed (vrms) as:
1/2 x m (vrms)2 = 3/2 kT
vrms= √(3KT/m)
Hence, the de Broglie wavelength of the nitrogen molecule is given as:
λ = h/(mvrms)= h/√(3mKT)
=(6.63 x 10⁻34) /√[3 x 28.0152 x (1.66 x 10⁻27) x (1.38 x 10⁻23) x 300
= 0.028 x 10⁻⁹ m
= 0.028 nm
Therefore, the de Broglie wavelength of the nitrogen molecule is 0.028 nm.
The momentum of a photon having energy (hv) is given as: p = hv/c = h/λ => λ = h/p Where, λ = Wavelength of the electromagnetic radiation c = Speed of light h = Planck's constant De Broglie wavelength of the photon is given as:λ = h/mv But , p=mv, therefore λ = h/p Where, m = Mass of the photonRead more
The momentum of a photon having energy (hv) is given as:
p = hv/c = h/λ
=> λ = h/p
Where, λ = Wavelength of the electromagnetic radiation
c = Speed of light
h = Planck’s constant
De Broglie wavelength of the photon is given as:λ = h/mv
But , p=mv, therefore
λ = h/p
Where, m = Mass of the photon v = Velocity of the photon
Hence, it can be inferred from equations (i) and (ii) that the wavelength of the electromagnetic radiation is equal to the de Broglie wavelength of the photon.
Ans (a). De Broglie wavelength of the neutron, λ = 1.40 x 10⁻10 m Mass of a neutron, mn = 1.66 x 10-27 kg Planck's constant, h = 6.6 x 10⁻34 Js Kinetic energy (K) and velocity (v) are related as : K = 1/2 mnv2 -----------------Eq-1 De Broglie wavelength (λ) and velocity (v) are related as : λ = h/mnRead more
Ans (a).
De Broglie wavelength of the neutron, λ = 1.40 x 10⁻10 m
Mass of a neutron, mn = 1.66 x 10-27 kg
Planck’s constant, h = 6.6 x 10⁻34 Js
Kinetic energy (K) and velocity (v) are related as :
K = 1/2 mnv2 —————–Eq-1
De Broglie wavelength (λ) and velocity (v) are related as :
λ = h/mnv—————–Eq-2
Using Eq-2 in Eq-1 .we get :
K = 1/2 x (mnh²)/(λ²m²n) = h²/2 λ²mn
=(6.6 x 10⁻34)² /2 (1.40 x 10⁻10)² (1.66 x 10-27)
= 6.75 x 10⁻²¹ J
= (6.75 x 10⁻²¹) /(1.6 x 10⁻¹⁹ )
= 4.219 x 10⁻² eV
Hence ,the kinetic energy of the neutron is 6.75 x 10⁻²¹ J or 4.219 x 10⁻² eV
Ans (b).
Temperature of the neutron, T = 300 K
Boltzmann constant, k = 1.38 x 10-23 kg m2 s⁻2 K⁻1
Average kinetic energy of the neutron:
K’ = 3/2 k T = 3/2 x 1.38 x 10-23 x 300 = 6.21 x 10-21 J
The relation for the de Broglie wavelength is given as:
λ’ =h/ √ (2K’mn)
Where, mn = 1.66 x 10-27 kg, h = 6.6 x 10-34 Js and K‘ = 6.75 x 10-21 J.
Therefore,
λ’ = (6.6 x 10-34)/√[2 x (6.21 x 10-21) x (1.66 x 10-27) ]
= 1.46 x 10–10 m = 0.146 nm
Therefore, the de Broglie wavelength of the neutron is 0.146 nm.
Wavelength of an electron (λe) and a photon (λp), λe = λp = 1 nm = 1 x 10⁻9 m Planck's constant, h = 6.63 x 10-34 Js Ans (a). The momentum of an elementary particle is given by de Broglie relation: λ = h/p => p = h/λ It is clear that momentum depends only on the wavelength of the particle. SincRead more
Wavelength of an electron (λe) and a photon (λp), λe = λp = 1 nm = 1 x 10⁻9 m
Planck’s constant, h = 6.63 x 10-34 Js
Ans (a).
The momentum of an elementary particle is given by de Broglie relation:
λ = h/p => p = h/λ
It is clear that momentum depends only on the wavelength of the particle. Since the wavelength of an electron and photon are equal, both have an equal momentum.
Therefore , p = 6.63 x 10⁻34 / ( 1 x 10⁻9)
= 6.63 x 10⁻²⁵ kg ms⁻¹
Ans (b).
The energy of a photon is given by the relation:
E = hc/λ
Where, Speed of light, c = 3 x 108 m/s ,
Therefore, E = (6.63 x 10⁻34 ) x (3 x 108 /(1 x 10⁻9) (1.6 x 10⁻¹⁹ )
=1243.1 eV = 1.243 kev
Therefore, the energy of the photon is 1.243 keV.
Ans (c).
The kinetic energy (K) of an electron having momentum p, is given by the relation:
K = 1/2 p²/m
Where,
m = Mass of the electron = 9.1 x 10-31 kg
p = 6.63 x 10-25 kg m s-1
Therefore , K = 1/2 ( 6.63 x 10-25 )² /( 9.1 x 10-31) = 2.415 x 10⁻¹⁹ J
= (2.415 x 10⁻¹⁹) /(1.6 x 10⁻¹⁹) = 1.51 eV
Hence, the kinetic energy of the electron is 1.51 eV.
Ans (a). Mass of the bullet, m = 0.040 kg Speed of the bullet, v = 1.0 km/s = 1000 m/s Planck’s constant, h = 6.6 x 10⁻34 Js De Broglie wavelength of the bullet is given by the relation: λ=h/mv =( 6.6 x 10⁻34)/(0.040) (1000) = 1.65 x 10 ⁻³⁵ m Ans (b). Mass of the ball, m = 0.060 kg Speed of the ballRead more
Ans (a).
Mass of the bullet, m = 0.040 kg
Speed of the bullet, v = 1.0 km/s = 1000 m/s
Planck’s constant, h = 6.6 x 10⁻34 Js
De Broglie wavelength of the bullet is given by the relation:
λ=h/mv
=( 6.6 x 10⁻34)/(0.040) (1000) = 1.65 x 10 ⁻³⁵ m
Ans (b).
Mass of the ball, m = 0.060 kg
Speed of the ball, v = 1.0 m/s
De Broglie wavelength of the ball is given by the relation:
λ=h/mv
=( 6.6 x 10⁻34)/(0.060) (1) = 1.1 x 10 ⁻³² m
Ans (c).
Mass of the dust particle, m = 1 x 10-9 kg
Speed of the dust particle, v = 2.2 m/s
De Broglie wavelength of the dust particle is given by the relation:
Wavelength of light of a sodium line, λ = 589 nm = 589 x 10-9 m Mass of an electron, me= 9.1 x 10-31 kg Mass of a neutron, mn= 1.66 x 10-27 kg Planck's constant, h = 6.6 x 10-34 Js Ans (a). For the kinetic energy K, of an electron accelerating with a velocity v, we have the relation: We have the relRead more
Wavelength of light of a sodium line, λ = 589 nm = 589 x 10–9 m
Mass of an electron, me= 9.1 x 10–31 kg
Mass of a neutron, mn= 1.66 x 10–27 kg
Planck’s constant, h = 6.6 x 10-34 Js
Ans (a).
For the kinetic energy K, of an electron accelerating with a velocity v, we have the relation:
We have the relation for de Broglie wavelength as:
K = 1/2 mev²—————-Eq-1
We have the relation for de Broglie wavelength as :
λ = h /mev
Therefore, v² =h²/λ²m²e——————— Eq-2 Substituting equation (2) in equation (1), we get the relation:
K = 1/2 meh²/2λ²m²e =h²/2λ²me—————-Eq-3
= (6.6 x 10-34)²/2 (589 x 10–9)² (9.1 x 10–31 )
≈ 6.9 x 10-25 J
=(6.9 x 10-25)/(1.6 x 10⁻¹⁹ ) =4.31 x 10 ⁻⁶ eV = 4.31 μ eV
Hence, the kinetic energy of the electron is 6.9 x 10-25 J or 4.31 μeV.
Ans (b).
Using equation (3), we can write the relation for the kinetic energy of the neutron as:
K= h²/2 λ²mn
= (6.6 x 10-34)² /2 (589 x 10–9)² (1.66 x 10–27 )
=3.78 x 10⁻²⁸ J
=( 3.78 x 10⁻²⁸ )/(1.6 x 10⁻¹⁹ )
= 2.36 x 10⁻⁹ eV = 2.36 neV
Hence, the kinetic energy of the neutron is 3.78 x 10⁻28 J or 2.36 neV.
(a) An X-ray tube produces a continuous spectrum of radiation with its short wavelength end at 0.45 Å. What is the maximum energy of a photon in the radiation? (b) From your answer to (a), guess what order of accelerating voltage (for electrons) is required in such a tube?
Ans (a). Wavelength produced by an X-ray tube, λ = 0.45 Aº = 0.45 x 10-10 m Planck’s constant, h = 6.626 x 10⁻34Js Speed of light, c = 3 x 108 m/s The maximum energy of a photon is given as: E = hc/λ = (6.626 x 10⁻34) ( 3 x 108 ) /(0.45 x 10-10) (1.6 x 10¹⁹ ) = 27.6x10³ eV = 27.6 keV Therefore, theRead more
Ans (a).
Wavelength produced by an X-ray tube, λ = 0.45 Aº = 0.45 x 10-10 m
Planck’s constant, h = 6.626 x 10⁻34Js
Speed of light, c = 3 x 108 m/s
The maximum energy of a photon is given as:
E = hc/λ = (6.626 x 10⁻34) ( 3 x 108 ) /(0.45 x 10-10) (1.6 x 10¹⁹ )
= 27.6×10³ eV = 27.6 keV
Therefore, the maximum energy of an X-ray photon is 27.6 keV.
Ans (b).
Accelerating voltage provides energy to the electrons for producing X-rays. To get an X-ray of 27.6 keV, the incident electrons must possess at least 27.6 keV of kinetic electric energy. Hence, an accelerating voltage of the order of 30 keV is required for producing X-rays.
See lessAn electron gun with its collector at a potential of 100 V fires out electrons in a spherical bulb containing hydrogen gas at low pressure (∼10⁻² mm of Hg). A magnetic field of 2.83 × 10⁻⁴ T curves the path of the electrons in a circular orbit of radius 12.0 cm. (The path can be viewed because the gas ions in the path focus the beam by attracting electrons, and emitting light by electron capture; this method is known as the ‘fine beam tube’ method.) Determine e/m from the data
Potential of an anode, V = 100 V Magnetic field experienced by the electrons, B = 2.83 x 10-4 T Radius of the circular orbit r = 12.0 cm = 12.0 x 10-2 m Mass of each electron = m and charge on each electron = e Velocity of each electron = v The energy of each electron is equal to its kinetic energy,Read more
Potential of an anode, V = 100 V
Magnetic field experienced by the electrons, B = 2.83 x 10-4 T
Radius of the circular orbit r = 12.0 cm = 12.0 x 10-2 m
Mass of each electron = m and
charge on each electron = e
Velocity of each electron = v
The energy of each electron is equal to its kinetic energy, i.e.,
1/2 mv² = eV
v²= 2eV/m————- Eq-1
It is the magnetic field, due to its bending nature, that provides the centripetal force (F = mv²/r) Centripetal force = Magnetic force
mv²/r = evB
eB = mv/r
v = eBr/m ——Eq-2
Putting the value of v in equation (1), we get:
2eV/m = e²B²r²/m²
e/m = 2V/B²r²
= 2 x 100 / (2.83 x 10-4 ) x (12.0 x 10-2) = 1.73 x 10¹¹Ckg-1
Therefore, the specific charge ratio (e/m) is 1.73 x 10¹¹ Ckg-1.
See less(a) A monoenergetic electron beam with electron speed of 5.20 × 10⁶ m s⁻¹ is subject to a magnetic field of 1.30 × 10⁻⁴ T normal to the beam velocity. What is the radius of the circle traced by the beam, given e/m for electron equals 1.76 × 10¹¹C kg⁻¹. (b) Is the formula you employ in (a) valid for calculating radius of the path of a 20 MeV electron beam? If not, in what way is it modified? [Note: Exercises 11.20(b) and 11.21(b) take you to relativistic mechanics which is beyond the scope of this book. They have been inserted here simply to emphasise the point that the formulas you use in part (a) of the exercises are not valid at very high speeds or energies. See answers at the end to know what ‘very high speed or energy’ means.]
Ans (a). Speed of an electron, v = 5.20 x 106 m/s Magnetic field experienced by the electron, B = 1.30 x 10-4 T Specific charge of an electron, e/m = 1.76 x 1011 C kg-1 Where, e = Charge on the electron = 1.6 x 10-19 C m = Mass of the electron = 9.1 x 10-31 kg-1 The force exerted on the electron isRead more
Ans (a).
Speed of an electron, v = 5.20 x 106 m/s
Magnetic field experienced by the electron, B = 1.30 x 10–4 T
Specific charge of an electron, e/m = 1.76 x 1011 C kg-1
Where, e = Charge on the electron = 1.6 x 10-19 C
m = Mass of the electron = 9.1 x 10–31 kg-1
The force exerted on the electron is given as:
F = e|v x B|
= evB sin 0
0 = Angle between the magnetic field and the beam velocity.
The magnetic field is normal to the direction of beam.
Therefore ,0 = 90°
F=evB …(1)
The beam traces a circular path of radius, r. It is the magnetic field, due to its bending nature, that provides the centripetal force (F = mv²/r) for the beam.
Hence, equation (1) reduces to:
evB = mv²/r
Therefore , r = mv/eB = v/(e/m)B
= (5.20 x 106)/(1.76 x 1011) x ( 1.30 x 10–4 ) =0.227 m =22.7 cm
Therefore, the radius of the circular path is 22.7 cm.
Ans (b).
Energy of the electron beam, E = 20 MeV = 20 x 106 x 1.6 x 10⁻19 J
The energy of the electron is given as:
E = 1/2 mv²
Therefore , v = √(2E/m)
= √ (2 x 20 x 106 x 1.6 x 10⁻19 )/(9.1 x 10–31) =2.652 x 10⁹ m/s
This result is incorrect because nothing can move faster than light. In the above formula, the expression (mv2/2) for energy can only be used in the non-relativistic limit, i.e., for v « c.
When very high speeds are concerned, the relativistic domain comes into consideration.
In the relativistic domain, mass is given as:
m = m0 √[1 – v²/c² ]
Where,
m0 = Mass of the particle at rest
Hence, the radius of the circular path is given as:
r = mv/eB = m0v/eB √[(c² – v²)/c² ]
See less(a) Estimate the speed with which electrons emitted from a heated emitter of an evacuated tube impinge on the collector maintained at a potential difference of 500 V with respect to the emitter. Ignore the small initial speeds of the electrons. The specific charge of the electron, i.e., its e/m is given to be 1.76 × 10¹¹ C kg⁻¹. (b) Use the same formula you employ in (a) to obtain electron speed for an collector potential of 10 MV. Do you see what is wrong ? In what way is the formula to be modified?
Ans (a). Potential difference across the evacuated tube, V = 500 V Specific charge of an electron, e/m = 1.76 x 1011 C kg-1 The speed of each emitted electron is given by the relation for kinetic energy as: KE = 1/2 mv²=eV Therefore ,v =√(2eV/m) = √(2V e/m) = √(2x 500 x 1.76 x 10¹¹) =1.327 x 107 m/sRead more
Ans (a).
Potential difference across the evacuated tube, V = 500 V
Specific charge of an electron, e/m = 1.76 x 1011 C kg-1
The speed of each emitted electron is given by the relation for kinetic energy as:
KE = 1/2 mv²=eV
Therefore ,v =√(2eV/m) = √(2V e/m)
= √(2x 500 x 1.76 x 10¹¹) =1.327 x 107 m/s
Therefore, the speed of each emitted electron is 1.327 x 107 m/s.
Ans (b).
Potential of the anode, V = 10 MV = 10 x 106 V
The speed of each electron is given as:
v = √ (2V e/m))
= √ (2 x 107 x 1.76x 10¹¹)- = 1.88 x 109 m/s
This result is wrong because nothing can move faster than light. In the above formula, the expression (mv2/2) for energy can only be used in the non-relativistic limit, i.e., for v « c.
For very high speed problems, relativistic equations must be considered for solving them. In the relativistic limit, the total energy is given as: E = me2
Where, m = Relativistic mass = mo √ (1 – v2/c2)
mo = Mass of the particle at rest Kinetic energy is given as: K = mc2 – moc2
See lessWhat is the de Broglie wavelength of a nitrogen molecule in air at 300 K? Assume that the molecule is moving with the root-mean-square speed of molecules at this temperature. (Atomic mass of nitrogen = 14.0076 u)
Temperature of the nitrogen molecule, T = 300 K Atomic mass of nitrogen = 14.0076 u Hence, mass of the nitrogen molecule, m = 2 x 14.0076 = 28.0152 u But 1 u = 1.66 x 10⁻27 kg Therefore, m = 28.0152 x 1.66 x 10⁻27 kg Planck’s constant, h = 6.63 x 10⁻34 Js Boltzmann constant, k = 1.38 x 10⁻23J K⁻1 WeRead more
Temperature of the nitrogen molecule, T = 300 K
Atomic mass of nitrogen = 14.0076 u
Hence, mass of the nitrogen molecule, m = 2 x 14.0076 = 28.0152 u
But 1 u = 1.66 x 10⁻27 kg
Therefore, m = 28.0152 x 1.66 x 10⁻27 kg
Planck’s constant, h = 6.63 x 10⁻34 Js
Boltzmann constant, k = 1.38 x 10⁻23J K⁻1
We have the expression that relates mean kinetic energy (3KT/2 )of the nitrogen molecule with the root mean square speed (vrms) as:
1/2 x m (vrms)2 = 3/2 kT
vrms= √(3KT/m)
Hence, the de Broglie wavelength of the nitrogen molecule is given as:
λ = h/(mvrms)= h/√(3mKT)
=(6.63 x 10⁻34) /√[3 x 28.0152 x (1.66 x 10⁻27) x (1.38 x 10⁻23) x 300
= 0.028 x 10⁻⁹ m
= 0.028 nm
Therefore, the de Broglie wavelength of the nitrogen molecule is 0.028 nm.
See lessShow that the wavelength of electromagnetic radiation is equal to the de Broglie wavelength of its quantum (photon).
The momentum of a photon having energy (hv) is given as: p = hv/c = h/λ => λ = h/p Where, λ = Wavelength of the electromagnetic radiation c = Speed of light h = Planck's constant De Broglie wavelength of the photon is given as:λ = h/mv But , p=mv, therefore λ = h/p Where, m = Mass of the photonRead more
The momentum of a photon having energy (hv) is given as:
p = hv/c = h/λ
=> λ = h/p
Where, λ = Wavelength of the electromagnetic radiation
c = Speed of light
h = Planck’s constant
De Broglie wavelength of the photon is given as:λ = h/mv
But , p=mv, therefore
λ = h/p
Where, m = Mass of the photon v = Velocity of the photon
Hence, it can be inferred from equations (i) and (ii) that the wavelength of the electromagnetic radiation is equal to the de Broglie wavelength of the photon.
See less(a) For what kinetic energy of a neutron will the associated de Broglie wavelength be 1.40 × 10⁻¹⁰ m? (b) Also find the de Broglie wavelength of a neutron, in thermal equilibrium with matter, having an average kinetic energy of (3/2) k T at 300 K
Ans (a). De Broglie wavelength of the neutron, λ = 1.40 x 10⁻10 m Mass of a neutron, mn = 1.66 x 10-27 kg Planck's constant, h = 6.6 x 10⁻34 Js Kinetic energy (K) and velocity (v) are related as : K = 1/2 mnv2 -----------------Eq-1 De Broglie wavelength (λ) and velocity (v) are related as : λ = h/mnRead more
Ans (a).
De Broglie wavelength of the neutron, λ = 1.40 x 10⁻10 m
Mass of a neutron, mn = 1.66 x 10-27 kg
Planck’s constant, h = 6.6 x 10⁻34 Js
Kinetic energy (K) and velocity (v) are related as :
K = 1/2 mnv2 —————–Eq-1
De Broglie wavelength (λ) and velocity (v) are related as :
λ = h/mnv—————–Eq-2
Using Eq-2 in Eq-1 .we get :
K = 1/2 x (mnh²)/(λ²m²n) = h²/2 λ²mn
=(6.6 x 10⁻34)² /2 (1.40 x 10⁻10)² (1.66 x 10-27)
= 6.75 x 10⁻²¹ J
= (6.75 x 10⁻²¹) /(1.6 x 10⁻¹⁹ )
= 4.219 x 10⁻² eV
Hence ,the kinetic energy of the neutron is 6.75 x 10⁻²¹ J or 4.219 x 10⁻² eV
Ans (b).
Temperature of the neutron, T = 300 K
Boltzmann constant, k = 1.38 x 10-23 kg m2 s⁻2 K⁻1
Average kinetic energy of the neutron:
K’ = 3/2 k T = 3/2 x 1.38 x 10-23 x 300 = 6.21 x 10-21 J
The relation for the de Broglie wavelength is given as:
λ’ =h/ √ (2K’mn)
Where, mn = 1.66 x 10-27 kg, h = 6.6 x 10-34 Js and K‘ = 6.75 x 10-21 J.
Therefore,
λ’ = (6.6 x 10-34)/√[2 x (6.21 x 10-21) x (1.66 x 10-27) ]
= 1.46 x 10–10 m = 0.146 nm
Therefore, the de Broglie wavelength of the neutron is 0.146 nm.
See lessAn electron and a photon each have a wavelength of 1.00 nm. Find (a) their momenta, (b) the energy of the photon, and (c) the kinetic energy of electron
Wavelength of an electron (λe) and a photon (λp), λe = λp = 1 nm = 1 x 10⁻9 m Planck's constant, h = 6.63 x 10-34 Js Ans (a). The momentum of an elementary particle is given by de Broglie relation: λ = h/p => p = h/λ It is clear that momentum depends only on the wavelength of the particle. SincRead more
Wavelength of an electron (λe) and a photon (λp), λe = λp = 1 nm = 1 x 10⁻9 m
Planck’s constant, h = 6.63 x 10-34 Js
Ans (a).
The momentum of an elementary particle is given by de Broglie relation:
λ = h/p => p = h/λ
It is clear that momentum depends only on the wavelength of the particle. Since the wavelength of an electron and photon are equal, both have an equal momentum.
Therefore , p = 6.63 x 10⁻34 / ( 1 x 10⁻9)
= 6.63 x 10⁻²⁵ kg ms⁻¹
Ans (b).
The energy of a photon is given by the relation:
E = hc/λ
Where, Speed of light, c = 3 x 108 m/s ,
Therefore, E = (6.63 x 10⁻34 ) x (3 x 108 /(1 x 10⁻9) (1.6 x 10⁻¹⁹ )
=1243.1 eV = 1.243 kev
Therefore, the energy of the photon is 1.243 keV.
Ans (c).
The kinetic energy (K) of an electron having momentum p, is given by the relation:
K = 1/2 p²/m
Where,
m = Mass of the electron = 9.1 x 10-31 kg
p = 6.63 x 10-25 kg m s-1
Therefore , K = 1/2 ( 6.63 x 10-25 )² /( 9.1 x 10-31) = 2.415 x 10⁻¹⁹ J
= (2.415 x 10⁻¹⁹) /(1.6 x 10⁻¹⁹) = 1.51 eV
Hence, the kinetic energy of the electron is 1.51 eV.
See lessWhat is the de Broglie wavelength of (a) a bullet of mass 0.040 kg travelling at the speed of 1.0 km/s, (b) a ball of mass 0.060 kg moving at a speed of 1.0 m/s, and (c) a dust particle of mass 1.0 × 10⁻⁹ kg drifting with a speed of 2.2 m/s?
Ans (a). Mass of the bullet, m = 0.040 kg Speed of the bullet, v = 1.0 km/s = 1000 m/s Planck’s constant, h = 6.6 x 10⁻34 Js De Broglie wavelength of the bullet is given by the relation: λ=h/mv =( 6.6 x 10⁻34)/(0.040) (1000) = 1.65 x 10 ⁻³⁵ m Ans (b). Mass of the ball, m = 0.060 kg Speed of the ballRead more
Ans (a).
Mass of the bullet, m = 0.040 kg
Speed of the bullet, v = 1.0 km/s = 1000 m/s
Planck’s constant, h = 6.6 x 10⁻34 Js
De Broglie wavelength of the bullet is given by the relation:
λ=h/mv
=( 6.6 x 10⁻34)/(0.040) (1000) = 1.65 x 10 ⁻³⁵ m
Ans (b).
Mass of the ball, m = 0.060 kg
Speed of the ball, v = 1.0 m/s
De Broglie wavelength of the ball is given by the relation:
λ=h/mv
=( 6.6 x 10⁻34)/(0.060) (1) = 1.1 x 10 ⁻³² m
Ans (c).
Mass of the dust particle, m = 1 x 10-9 kg
Speed of the dust particle, v = 2.2 m/s
De Broglie wavelength of the dust particle is given by the relation:
λ=h/mv
=( 6.6 x 10⁻34)/(2.2) (1 x 10-9) = 3.0 x 10 ⁻²⁵ m
See lessThe wavelength of light from the spectral emission line of sodium is 589 nm. Find the kinetic energy at which (a) an electron, and (b) a neutron, would have the same de Broglie wavelength.
Wavelength of light of a sodium line, λ = 589 nm = 589 x 10-9 m Mass of an electron, me= 9.1 x 10-31 kg Mass of a neutron, mn= 1.66 x 10-27 kg Planck's constant, h = 6.6 x 10-34 Js Ans (a). For the kinetic energy K, of an electron accelerating with a velocity v, we have the relation: We have the relRead more
Wavelength of light of a sodium line, λ = 589 nm = 589 x 10–9 m
Mass of an electron, me= 9.1 x 10–31 kg
Mass of a neutron, mn= 1.66 x 10–27 kg
Planck’s constant, h = 6.6 x 10-34 Js
Ans (a).
For the kinetic energy K, of an electron accelerating with a velocity v, we have the relation:
We have the relation for de Broglie wavelength as:
K = 1/2 mev²—————-Eq-1
We have the relation for de Broglie wavelength as :
λ = h /mev
Therefore, v² =h²/λ²m²e——————— Eq-2
Substituting equation (2) in equation (1), we get the relation:
K = 1/2 meh²/2λ²m²e =h²/2λ²me—————-Eq-3
= (6.6 x 10-34)²/2 (589 x 10–9)² (9.1 x 10–31 )
≈ 6.9 x 10-25 J
=(6.9 x 10-25)/(1.6 x 10⁻¹⁹ ) =4.31 x 10 ⁻⁶ eV = 4.31 μ eV
Hence, the kinetic energy of the electron is 6.9 x 10-25 J or 4.31 μeV.
Ans (b).
Using equation (3), we can write the relation for the kinetic energy of the neutron as:
K= h²/2 λ²mn
= (6.6 x 10-34)² /2 (589 x 10–9)² (1.66 x 10–27 )
=3.78 x 10⁻²⁸ J
=( 3.78 x 10⁻²⁸ )/(1.6 x 10⁻¹⁹ )
= 2.36 x 10⁻⁹ eV = 2.36 neV
Hence, the kinetic energy of the neutron is 3.78 x 10⁻28 J or 2.36 neV.
See less