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.
Ans (a). The initial velocity of the particle is either parallel or anti-parallel to the magnetic field. Hence, it travels along a straight path without suffering any deflection in the field. Ans (b). Yes, the final speed of the charged particle will be equal to its initial speed. This is because maRead more
Ans (a).
The initial velocity of the particle is either parallel or anti-parallel to the magnetic field. Hence, it travels along a straight path without suffering any deflection in the field.
Ans (b).
Yes, the final speed of the charged particle will be equal to its initial speed. This is because magnetic force can change the direction of velocity, but not its magnitude.
Ans (c).
An electron travelling from West to East enters a chamber having a uniform electrostatic field in the North-South direction. This moving electron can remain un-deflected if the electric force acting on it is equal and opposite of magnetic field. Magnetic force is directed towards the South. According to Fleming’s left hand rule, magnetic field should be applied in a vertically downward direction.
Inner radius of the toroid, r₁ = 25 cm = 0.25 m Outer radius of the toroid, r₂ = 26 cm = 0.26 m Number of turns on the coil, N = 3500 Current in the coil, I = 11 A Ans (a). Magnetic field outside a toroid is zero. It is non-zero only inside the core of a toroid, (b) Magnetic field inside the core oRead more
Inner radius of the toroid, r₁ = 25 cm = 0.25 m
Outer radius of the toroid, r₂ = 26 cm = 0.26 m
Number of turns on the coil, N = 3500
Current in the coil, I = 11 A
Ans (a).
Magnetic field outside a toroid is zero. It is non-zero only inside the core of a toroid, (b) Magnetic field inside the core of a toroid is given by the relation, B =
Where, μ0 = Permeability of free space = 4π x 10⁻⁷ TmA⁻¹
l = length of toroid = 2π (r₁ + r₂)/2=2 π (0.25 +0.26)=0.51 π
Therefore B = (4π x 10⁻⁷ x 3500 x 11 )/0.51 π = 3 .0 x 10⁻² T
Ans (c).
Magnetic field in the empty space surrounded by the toroid is zero.
The distance between electron-proton of a hydrogen atom, d = 0.53 A Charge on an electron, q₁ = -1.6 x 10-19 C Charge on a proton, q2 = +1.6 x 10-19 C Ans (a). Potential at infinity is zero. Potential energy of the system, p-e = Potential energy at infinity - Potential energy at distance, d = 0 - (qRead more
The distance between electron-proton of a hydrogen atom, d = 0.53 A
Charge on an electron, q₁ = -1.6 x 10-19 C Charge on a proton, q2 = +1.6 x 10-19 C
Ans (a).
Potential at infinity is zero.
Potential energy of the system, p-e
= Potential energy at infinity – Potential energy at distance, d
= 0 – (q₁-q₂)/4πε0 d
Where, ε0 is the permittivity of free space and
1/4πε0 =9x 10⁹ Nm²C⁻²
Therefore Potential Energy = 0- (9x 10⁹) x (1.6 x 10⁻¹⁹)² /(0.53 x 10¹⁰)
= -43.7x 10⁻¹⁹ J , (Since 1.6 x 10⁻¹⁹ J = 1 eV)
Therefore Potential energy =-43.7×10⁻¹⁹ = -(43.7 x 10⁻¹⁹)/(1.6 x 10⁻¹⁹) =-27.2 eV
Therefore, the potential energy of the system is -27.2 eV.
Ans (b).
Kinetic energy is half of the magnitude of potential energy.
Kinetic energy = 1/2 x(-27.2) = 13.6 eV
Total energy = 13.6 – 27.2 = 13.6 eV
Therefore, the minimum work required to free the electron is 13.6 eV.
Ans (c).
When zero of potential energy is taken, d = 1.06 A
Therefore Potential energy of the system = Potential energy at d₁- Potential energy at d
= (q₁q₂)/4πε0 d₁ – 27.2 eV
= (9 x 10⁹) x (1.6 x 10⁻¹⁹)² / ( 1.06 x 10⁻¹⁰) -27.2 eV
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 lessAnswer the following questions: (a) A magnetic field that varies in magnitude from point to point but has a constant direction (east to west) is set up in a chamber. A charged particle enters the chamber and travels undeflected along a straight path with constant speed. What can you say about the initial velocity of the particle? (b) A charged particle enters an environment of a strong and non-uniform magnetic field varying from point to point both in magnitude and direction, and comes out of it following a complicated trajectory. Would its final speed equal the initial speed if it suffered no collisions with the environment? (c) An electron travelling west to east enters a chamber having a uniform electrostatic field in north to south direction. Specify the direction in which a uniform magnetic field should be set up to prevent the electron from deflecting from its straight line path.
Ans (a). The initial velocity of the particle is either parallel or anti-parallel to the magnetic field. Hence, it travels along a straight path without suffering any deflection in the field. Ans (b). Yes, the final speed of the charged particle will be equal to its initial speed. This is because maRead more
Ans (a).
The initial velocity of the particle is either parallel or anti-parallel to the magnetic field. Hence, it travels along a straight path without suffering any deflection in the field.
Ans (b).
Yes, the final speed of the charged particle will be equal to its initial speed. This is because magnetic force can change the direction of velocity, but not its magnitude.
Ans (c).
An electron travelling from West to East enters a chamber having a uniform electrostatic field in the North-South direction. This moving electron can remain un-deflected if the electric force acting on it is equal and opposite of magnetic field. Magnetic force is directed towards the South. According to Fleming’s left hand rule, magnetic field should be applied in a vertically downward direction.
See lessA toroid has a core (non-ferromagnetic) of inner radius 25 cm and outer radius 26 cm, around which 3500 turns of a wire are wound. If the current in the wire is 11 A, what is the magnetic field (a) outside the toroid, (b) inside the core of the toroid, and (c) in the empty space surrounded by the toroid.
Inner radius of the toroid, r₁ = 25 cm = 0.25 m Outer radius of the toroid, r₂ = 26 cm = 0.26 m Number of turns on the coil, N = 3500 Current in the coil, I = 11 A Ans (a). Magnetic field outside a toroid is zero. It is non-zero only inside the core of a toroid, (b) Magnetic field inside the core oRead more
Inner radius of the toroid, r₁ = 25 cm = 0.25 m
Outer radius of the toroid, r₂ = 26 cm = 0.26 m
Number of turns on the coil, N = 3500
Current in the coil, I = 11 A
Ans (a).
Magnetic field outside a toroid is zero. It is non-zero only inside the core of a toroid, (b) Magnetic field inside the core of a toroid is given by the relation, B =
Where, μ0 = Permeability of free space = 4π x 10⁻⁷ TmA⁻¹
l = length of toroid = 2π (r₁ + r₂)/2=2 π (0.25 +0.26)=0.51 π
Therefore B = (4π x 10⁻⁷ x 3500 x 11 )/0.51 π = 3 .0 x 10⁻² T
Ans (c).
Magnetic field in the empty space surrounded by the toroid is zero.
See lessIn a hydrogen atom, the electron and proton are bound at a distance of about 0.53 Å:(a) Estimate the potential energy of the system in eV, taking the zero of the potential energy at infinite separation of the electron from proton. (b) What is the minimum work required to free the electron, given that its kinetic energy in the orbit is half the magnitude of potential energy obtained in (a)? (c) What are the answers to (a) and (b) above if the zero of potential energy is taken at 1.06 Å separation?
The distance between electron-proton of a hydrogen atom, d = 0.53 A Charge on an electron, q₁ = -1.6 x 10-19 C Charge on a proton, q2 = +1.6 x 10-19 C Ans (a). Potential at infinity is zero. Potential energy of the system, p-e = Potential energy at infinity - Potential energy at distance, d = 0 - (qRead more
The distance between electron-proton of a hydrogen atom, d = 0.53 A
Charge on an electron, q₁ = -1.6 x 10-19 C Charge on a proton, q2 = +1.6 x 10-19 C
Ans (a).
Potential at infinity is zero.
Potential energy of the system, p-e
= Potential energy at infinity – Potential energy at distance, d
= 0 – (q₁-q₂)/4πε0 d
Where, ε0 is the permittivity of free space and
1/4πε0 =9x 10⁹ Nm²C⁻²
Therefore Potential Energy = 0- (9x 10⁹) x (1.6 x 10⁻¹⁹)² /(0.53 x 10¹⁰)
= -43.7x 10⁻¹⁹ J , (Since 1.6 x 10⁻¹⁹ J = 1 eV)
Therefore Potential energy =-43.7×10⁻¹⁹ = -(43.7 x 10⁻¹⁹)/(1.6 x 10⁻¹⁹) =-27.2 eV
Therefore, the potential energy of the system is -27.2 eV.
Ans (b).
Kinetic energy is half of the magnitude of potential energy.
Kinetic energy = 1/2 x(-27.2) = 13.6 eV
Total energy = 13.6 – 27.2 = 13.6 eV
Therefore, the minimum work required to free the electron is 13.6 eV.
Ans (c).
When zero of potential energy is taken, d = 1.06 A
Therefore Potential energy of the system = Potential energy at d₁- Potential energy at d
= (q₁q₂)/4πε0 d₁ – 27.2 eV
= (9 x 10⁹) x (1.6 x 10⁻¹⁹)² / ( 1.06 x 10⁻¹⁰) -27.2 eV
=21.73 x 10⁻¹⁹ J – 27.2 eV =
=13.58 eV – 27.2 eV= – 13.6 eV
See less