Photoelectric cut-off voltage, Vo = 1.5 V The maximum kinetic energy of the emitted photoelectrons is given as: Ke=eV0 Where, e = Charge on an electron = 1.6 x 10-19 C Therefore, Ke = 1.6 x 10-19 x 1.5 = 2.4 x 10-19 J Therefore, the maximum kinetic energy of the photoelectrons emitted in the given eRead more
Photoelectric cut-off voltage, Vo = 1.5 V
The maximum kinetic energy of the emitted photoelectrons is given as:
Ke=eV0
Where, e = Charge on an electron = 1.6 x 10-19 C
Therefore, Ke = 1.6 x 10-19 x 1.5
= 2.4 x 10-19 J
Therefore, the maximum kinetic energy of the photoelectrons emitted in the given experiment is 2.4 x 10-19J.
Work function of caesium metal, φO = 2.14 eV and frequency of light, v = 6.0 x 1014 Hz Ans (a). The maximum kinetic energy is given by the photoelectric effect as: K = hv — φO Where, h = Planck’s constant = 6.626 x 10⁻34 Js Therefore ,K = (6.626 x 10⁻34) x ( 6.0 x 1014) /(1.6 x 10⁻¹⁹) -2.140 = 2.485Read more
Work function of caesium metal, φO = 2.14 eV and
frequency of light, v = 6.0 x 1014 Hz
Ans (a).
The maximum kinetic energy is given by the photoelectric effect as: K = hv — φO
Where, h = Planck’s constant = 6.626 x 10⁻34 Js
Therefore ,K = (6.626 x 10⁻34) x ( 6.0 x 1014) /(1.6 x 10⁻¹⁹) -2.140
= 2.485 -2.140 = 0.345 eV
Hence, the maximum kinetic energy of the emitted electrons is 0.345 eV.
Ans (b).
For stopping potential V0 we can write the equation for kinetic energy as:
K=e V0
Therefore, V0= K/e = (0.345 x 1.6 x10⁻¹⁹) /(1.6 x 10⁻¹⁹ )
Hence ,the stopping potential of the material is 0.345 V
Ans (c).
Maximum speed of the emitted photoelectrons = v
Hence, the relation for kinetic energy can be written as:
K = 1/2 x (mv²)
Where, m = Mass of an electron = 9.1 x 10-31 kg
v² =2K /m
= 2 x (0.345 x 1.6 x 10⁻¹⁹ )/(9.1 x 10-31) = 0.1104 x 10¹²
v=3.323 x 10⁵ m/s = 332.3 km/s
Hence, the maximum speed of the emitted photoelectrons is 332.3 km/s.
Magnetic moment of the bar magnet, M = 5.25 x 10⁻2 J T⁻1 Magnitude of earth's magnetic field at a place, H = 0.42 G = 0.42 x 10⁻4 T Ans (a). The magnetic field at a distance R from the centre of the magnet on the normal bisector is given by the relation: B₁ = μ0 /4π x [M/(R)³] Where, μ0 = PermeabiliRead more
Magnetic moment of the bar magnet, M = 5.25 x 10⁻2 J T⁻1
Magnitude of earth’s magnetic field at a place, H = 0.42 G = 0.42 x 10⁻4 T
Ans (a).
The magnetic field at a distance R from the centre of the magnet on the normal bisector is given by the relation:
B₁ = μ0 /4π x [M/(R)³]
Where,
μ0 = Permeability of free space = 4π x 10-7 Tm A-1
When the resultant field is inclined at 45° with earth’s field, B = H
Therefore , μ0 /4π x [M/(R)³] =H = 0.42 x 10⁻4
=> R³ = μ0M /( 0.42 x 10⁻4 x 4π )
= (4π x 10-7 x 5.25 x 10⁻2 )/ ( 0.42 x 10⁻4 x 4π ) = 12.5 x 10⁻⁵
Therefore R = 0.05m = 5 cm
Ans (b).
The magnetic field at a distance R1 from the centre of the magnet on its axis is given as:
B¹ = μ0 /4π x [2M/(R¹)³]
The resultant field is inclined at 45° with earth’s field.
Therefore ,
B¹ =H
μ0 /4π x [2M/(R¹)³] = H
(R¹)³ = μ0 2M/ 4πH
= (4π x 10-7 x 2 x 5.25 x 10⁻2)/ (4π x 0.42 x 10⁻4 )
The magnetic field on the axis of the magnet at a distance d₁= 14 cm, can be written as: B₁ = μ0 /4π x [2M/(d₁)³] =H -----------Eq 1 Where, M = Magnetic moment μ0= Permeability of free space, H = Horizontal component of the magnetic field at d₁. If the bar magnet is turned through 180°, then the nRead more
The magnetic field on the axis of the magnet at a distance d₁= 14 cm, can be written as:
B₁ = μ0 /4π x [2M/(d₁)³] =H ———–Eq 1
Where, M = Magnetic moment
μ0= Permeability of free space,
H = Horizontal component of the magnetic field at d₁.
If the bar magnet is turned through 180°, then the neutral point will lie on the equatorial line.
Hence, the magnetic field at a distance d₂, on the equatorial line of the magnet can be written as:
B₂ = μ0 /4π x [2M/(d₂)³] =H ———–Eq 2
Equating equations (1) and (2), we get:
2/(d₁)³= 1/(d₂)³
=> (d₂/d₁)³ = 1/2
=> d₂ = d₁ x (1/2)1/3
=> d₂ = 14 x 0.794 = 11.1 cm
The new null points will be located 11.1 cm on the normal bisector.
Earth's magnetic field at the given place, H = 0.36 G The magnetic field at a distance d, on the axis of the magnet is given as: B₁ = μ0 /4π x (2M/d³) = H ------------------Eq -1 Where, μ0 = Permeability of free space, M = Magnetic moment The magnetic field at the same distance d, on the equatorialRead more
Earth’s magnetic field at the given place, H = 0.36 G The magnetic field at a distance d, on the axis of the magnet is given as: B₁ = μ0 /4π x (2M/d³) = H ——————Eq -1
Where, μ0 = Permeability of free space, M = Magnetic moment
The magnetic field at the same distance d, on the equatorial line of the magnet is given as:
B₂ = μ0 /4π x (M/d³) = H/2 (Using Eq -1)
Total magnetic field ,B = B₁ + B₂
= 0.36 + 0.18 = 0.54 G
Hence, the magnetic field is 0.54 G in the direction of earth’s magnetic field.
The photoelectric cut-off voltage in a certain experiment is 1.5 V. What is the maximum kinetic energy of photoelectrons emitted?
Photoelectric cut-off voltage, Vo = 1.5 V The maximum kinetic energy of the emitted photoelectrons is given as: Ke=eV0 Where, e = Charge on an electron = 1.6 x 10-19 C Therefore, Ke = 1.6 x 10-19 x 1.5 = 2.4 x 10-19 J Therefore, the maximum kinetic energy of the photoelectrons emitted in the given eRead more
Photoelectric cut-off voltage, Vo = 1.5 V
The maximum kinetic energy of the emitted photoelectrons is given as:
Ke=eV0
Where, e = Charge on an electron = 1.6 x 10-19 C
Therefore, Ke = 1.6 x 10-19 x 1.5
= 2.4 x 10-19 J
Therefore, the maximum kinetic energy of the photoelectrons emitted in the given experiment is 2.4 x 10-19J.
See lessThe work function of caesium metal is 2.14 eV. When light of frequency 6 ×10¹⁴ Hz is incident on the metal surface, photoemission of electrons occurs. What is the (a) maximum kinetic energy of the emitted electrons, (b) Stopping potential, and (c) maximum speed of the emitted photoelectrons?
Work function of caesium metal, φO = 2.14 eV and frequency of light, v = 6.0 x 1014 Hz Ans (a). The maximum kinetic energy is given by the photoelectric effect as: K = hv — φO Where, h = Planck’s constant = 6.626 x 10⁻34 Js Therefore ,K = (6.626 x 10⁻34) x ( 6.0 x 1014) /(1.6 x 10⁻¹⁹) -2.140 = 2.485Read more
Work function of caesium metal, φO = 2.14 eV and
frequency of light, v = 6.0 x 1014 Hz
Ans (a).
The maximum kinetic energy is given by the photoelectric effect as: K = hv — φO
Where, h = Planck’s constant = 6.626 x 10⁻34 Js
Therefore ,K = (6.626 x 10⁻34) x ( 6.0 x 1014) /(1.6 x 10⁻¹⁹) -2.140
= 2.485 -2.140 = 0.345 eV
Hence, the maximum kinetic energy of the emitted electrons is 0.345 eV.
Ans (b).
For stopping potential V0 we can write the equation for kinetic energy as:
K=e V0
Therefore, V0= K/e = (0.345 x 1.6 x10⁻¹⁹) /(1.6 x 10⁻¹⁹ )
Hence ,the stopping potential of the material is 0.345 V
Ans (c).
Maximum speed of the emitted photoelectrons = v
Hence, the relation for kinetic energy can be written as:
K = 1/2 x (mv²)
Where, m = Mass of an electron = 9.1 x 10-31 kg
v² =2K /m
= 2 x (0.345 x 1.6 x 10⁻¹⁹ )/(9.1 x 10-31) = 0.1104 x 10¹²
v=3.323 x 10⁵ m/s = 332.3 km/s
Hence, the maximum speed of the emitted photoelectrons is 332.3 km/s.
See lessA short bar magnet of magnetic moment 5.25 × 10⁻² J T⁻¹ is placed with its axis perpendicular to the earth’s field direction. At what distance from the centre of the magnet, the resultant field is inclined at 45º with earth’s field on (a) its normal bisector and (b) its axis. Magnitude of the earth’s field at the place is given to be 0.42 G. Ignore the length of the magnet in comparison to the distances involved.
Magnetic moment of the bar magnet, M = 5.25 x 10⁻2 J T⁻1 Magnitude of earth's magnetic field at a place, H = 0.42 G = 0.42 x 10⁻4 T Ans (a). The magnetic field at a distance R from the centre of the magnet on the normal bisector is given by the relation: B₁ = μ0 /4π x [M/(R)³] Where, μ0 = PermeabiliRead more
Magnetic moment of the bar magnet, M = 5.25 x 10⁻2 J T⁻1
Magnitude of earth’s magnetic field at a place, H = 0.42 G = 0.42 x 10⁻4 T
Ans (a).
The magnetic field at a distance R from the centre of the magnet on the normal bisector is given by the relation:
B₁ = μ0 /4π x [M/(R)³]
Where,
μ0 = Permeability of free space = 4π x 10-7 Tm A-1
When the resultant field is inclined at 45° with earth’s field, B = H
Therefore , μ0 /4π x [M/(R)³] =H = 0.42 x 10⁻4
=> R³ = μ0M /( 0.42 x 10⁻4 x 4π )
= (4π x 10-7 x 5.25 x 10⁻2 )/ ( 0.42 x 10⁻4 x 4π ) = 12.5 x 10⁻⁵
Therefore R = 0.05m = 5 cm
Ans (b).
The magnetic field at a distance R1 from the centre of the magnet on its axis is given as:
B¹ = μ0 /4π x [2M/(R¹)³]
The resultant field is inclined at 45° with earth’s field.
Therefore ,
B¹ =H
μ0 /4π x [2M/(R¹)³] = H
(R¹)³ = μ0 2M/ 4πH
= (4π x 10-7 x 2 x 5.25 x 10⁻2)/ (4π x 0.42 x 10⁻4 )
= 25 x 10⁻⁵
Therefore , R¹ = 0.063 m = 6.3 cm
See lessIf the bar magnet in exercise 5.13 is turned around by 180º, where will the new null points be located?
The magnetic field on the axis of the magnet at a distance d₁= 14 cm, can be written as: B₁ = μ0 /4π x [2M/(d₁)³] =H -----------Eq 1 Where, M = Magnetic moment μ0= Permeability of free space, H = Horizontal component of the magnetic field at d₁. If the bar magnet is turned through 180°, then the nRead more
The magnetic field on the axis of the magnet at a distance d₁= 14 cm, can be written as:
B₁ = μ0 /4π x [2M/(d₁)³] =H ———–Eq 1
Where, M = Magnetic moment
μ0= Permeability of free space,
H = Horizontal component of the magnetic field at d₁.
If the bar magnet is turned through 180°, then the neutral point will lie on the equatorial line.
Hence, the magnetic field at a distance d₂, on the equatorial line of the magnet can be written as:
B₂ = μ0 /4π x [2M/(d₂)³] =H ———–Eq 2
Equating equations (1) and (2), we get:
2/(d₁)³= 1/(d₂)³
=> (d₂/d₁)³ = 1/2
=> d₂ = d₁ x (1/2)1/3
=> d₂ = 14 x 0.794 = 11.1 cm
The new null points will be located 11.1 cm on the normal bisector.
See lessA short bar magnet placed in a horizontal plane has its axis aligned along the magnetic north-south direction. Null points are found on the axis of the magnet at 14 cm from the centre of the magnet. The earth’s magnetic field at the place is 0.36 G and the angle of dip is zero. What is the total magnetic field on the normal bisector of the magnet at the same distance as the null–point (i.e., 14 cm) from the centre of the magnet? (At null points, field due to a magnet is equal and opposite to the horizontal component of earth’s magnetic field.)
Earth's magnetic field at the given place, H = 0.36 G The magnetic field at a distance d, on the axis of the magnet is given as: B₁ = μ0 /4π x (2M/d³) = H ------------------Eq -1 Where, μ0 = Permeability of free space, M = Magnetic moment The magnetic field at the same distance d, on the equatorialRead more
Earth’s magnetic field at the given place, H = 0.36 G
The magnetic field at a distance d, on the axis of the magnet is given as:
B₁ = μ0 /4π x (2M/d³) = H ——————Eq -1
Where, μ0 = Permeability of free space, M = Magnetic moment
The magnetic field at the same distance d, on the equatorial line of the magnet is given as:
B₂ = μ0 /4π x (M/d³) = H/2 (Using Eq -1)
Total magnetic field ,B = B₁ + B₂
= 0.36 + 0.18 = 0.54 G
Hence, the magnetic field is 0.54 G in the direction of earth’s magnetic field.
See less