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.
(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 less