Kinetic energy of the electron, Ek = 120 eV Planck's constant, h = 6.6 x 10⁻34Js Mass of an electron, m = 9.1 x 10⁻31 kg Charge on an electron, e = 1.6 x 10⁻19 C Ans (a). For the electron, we can write the relation for kinetic energy as: Ek = 1/2 mv² Where, v = Speed of the electron Therefore, v =√(Read more
Kinetic energy of the electron, Ek = 120 eV
Planck’s constant, h = 6.6 x 10⁻34Js
Mass of an electron, m = 9.1 x 10⁻31 kg
Charge on an electron, e = 1.6 x 10⁻19 C Ans (a).
For the electron, we can write the relation for kinetic energy as:
Ek = 1/2 mv²
Where, v = Speed of the electron
Therefore, v =√(2Ek/m) =√( 2 x 120 x 1.6 x 10 x 120 )/(9.1 x 10⁻31 )
=√ (42.198 x 10¹²)= 6.496 x 10⁶ m/s
Momentum of the electron, p = mv = 9.1 x 10⁻31 x 6.496 x 10⁶
=5.91 x 10²⁴ kg ms⁻1
Therefore, the momentum of the electron is 5.91 x 10-24 kg m s-1.
Ans (b).
Speed of the electron, v = 6.496 x 106 m/s
Ans (c).
De Broglie wavelength of an electron having a momentum p, is given as:
λ = h/p = ( 6.6 x 10⁻34) /( 5.91 x 10-24) = 1.116 x 10⁻¹⁰m
= 0.112 nm
Therefore, the de Broglie wavelength of the electron is 0.112 nm.
Potential difference, V = 56 V Planck's constant, h = 6.6 x 10⁻34 Js Mass of an electron, m = 9.1 x 10⁻31 kg Charge on an electron, e = 1.6 x 10⁻19 C Ans (a). At equilibrium, the kinetic energy of each electron is equal to the accelerating potential, i.e., we can write the relation for velocity (v)Read more
Potential difference, V = 56 V
Planck’s constant, h = 6.6 x 10⁻34 Js
Mass of an electron, m = 9.1 x 10⁻31 kg
Charge on an electron, e = 1.6 x 10⁻19 C
Ans (a).
At equilibrium, the kinetic energy of each electron is equal to the accelerating potential, i.e., we can write the relation for velocity (v) of each electron as:
-1/2mv2=eV
v2 =2eV/m
Therefore ,v = √ (2 x 1.6 x10⁻19 x 56)/ (9.1 x 10⁻31)
= √(19.69 x 10¹² )=4.44 x 10⁶ m/s
The momentum of each accelerated electron is given as:
p = mv = 9.1 x 10-31 x 4.44 x 106 = 4.04 x 10-24 kg m s⁻1Therefore, the momentum of each electron is 4.04 x 10-24 kg m s-1.
Ans (b).
De Broglie wavelength of an electron accelerating through a potential V, is given by the relation:
λ = 12.27 Aº/√V
= 12.27 x 10⁻¹⁰ /√56
= 0.1639 nm
Therefore, the de Broglie wavelength of each electron is 0.1639 nm.
Wavelength of light produced by the argon laser, λ=488nm = 488 x 10-9m Stopping potential of the photoelectrons, Vo = 0.38 V We know that 1 eV = 1.6 x 10-19 J, therefore Vo =0.38/1.6 x 10-19 eV Planck’s constant, h = 6.6 x 10-34 Js Charge on an electron, e = 1.6 x 10-19 C Speed of light, c = 3 x 10⁸Read more
Wavelength of light produced by the argon laser,
λ=488nm = 488 x 10-9m
Stopping potential of the photoelectrons,
Vo = 0.38 V
We know that 1 eV = 1.6 x 10-19 J, therefore
Vo =0.38/1.6 x 10-19 eV
Planck’s constant, h = 6.6 x 10-34 Js
Charge on an electron, e = 1.6 x 10-19 C
Speed of light, c = 3 x 10⁸ m/s
From Einstein’s photoelectric effect, we have the relation involving the work function φo of the material of the emitter as:
eVo = hc/λ – φo
Therefore , φo = hc/λ -eVo
= (6.6 x 10-34) x (3 x 10⁸ ) /(1.6 x 10-19) ( 488 x 10-9) – (1.6 x 10-19) x 0.38/(1.6 x 10-19)
= 2.54 -0.38 = 2.16 eV
Therefore, the material with which the emitter is made has the work function of 2.16 eV.
Frequency of the incident photon, v =488 nm = 488 x 10-9 m Maximum speed of the electrons, v = 6.0 x 105 m/s Planck’s constant, h = 6.626 x 10-34 Js Mass of an electron, m = 9.1 x 10-31 kg For threshold frequency vo, the relation for kinetic energy is written as: 1/2 (mv2 )=h(v-v0 ) v0 = v -mv2 /2hRead more
Frequency of the incident photon, v =488 nm = 488 x 10-9 m
Maximum speed of the electrons, v = 6.0 x 105 m/s
Planck’s constant, h = 6.626 x 10-34 Js
Mass of an electron, m = 9.1 x 10-31 kg
For threshold frequency vo, the relation for kinetic energy is written as:
1/2 (mv2 )=h(v-v0 )
v0 = v -mv2 /2h
= (7.21 x 1014 ) – (9.1 x 10-31) x (6.0 x 105)2 /2 x ( 6.626 x 10-34 )
= (7.21 x 1014 ) -2.472 x 1014
= 4.738 x 1014 Hz
Therefore, the threshold frequency for the photoemission of electrons is 4.738 x 1014 Hz.
No Work function of the metal, φO = 4.2 eV Charge on an electron, e = 1.6 x 10-19 C Planck’s constant, h = 6.626 x 10-34 Js Wavelength of the incident radiation, λ = 330 nm = 330 x 10⁻9 m Speed of light, c = 3 x 108 m/s The energy of the incident photon is given as: E = hc/λ = (6.626 x 10-34) x ( 3Read more
No
Work function of the metal, φO = 4.2 eV
Charge on an electron, e = 1.6 x 10-19 C
Planck’s constant, h = 6.626 x 10–34 Js
Wavelength of the incident radiation, λ = 330 nm = 330 x 10⁻9 m
Speed of light, c = 3 x 108 m/s
The energy of the incident photon is given as:
E = hc/λ = (6.626 x 10–34) x ( 3 x 108) /(330 x 10⁻9)
=6.0 x 10⁻19J
=(6.0 x 10⁻19J)/(1.6 x 10⁻19 ) = 3.76 eV
It can be observed that the energy of the incident radiation is less than the work function of the metal. Hence, no photoelectric emission will take place.
What is the (a) momentum, (b) speed, and (c) de Broglie wavelength of an electron with kinetic energy of 120 eV.
Kinetic energy of the electron, Ek = 120 eV Planck's constant, h = 6.6 x 10⁻34Js Mass of an electron, m = 9.1 x 10⁻31 kg Charge on an electron, e = 1.6 x 10⁻19 C Ans (a). For the electron, we can write the relation for kinetic energy as: Ek = 1/2 mv² Where, v = Speed of the electron Therefore, v =√(Read more
Kinetic energy of the electron, Ek = 120 eV
Planck’s constant, h = 6.6 x 10⁻34Js
Mass of an electron, m = 9.1 x 10⁻31 kg
Charge on an electron, e = 1.6 x 10⁻19 C
Ans (a).
For the electron, we can write the relation for kinetic energy as:
Ek = 1/2 mv²
Where, v = Speed of the electron
Therefore, v =√(2Ek/m) =√( 2 x 120 x 1.6 x 10 x 120 )/(9.1 x 10⁻31 )
=√ (42.198 x 10¹²)= 6.496 x 10⁶ m/s
Momentum of the electron, p = mv = 9.1 x 10⁻31 x 6.496 x 10⁶
=5.91 x 10²⁴ kg ms⁻1
Therefore, the momentum of the electron is 5.91 x 10-24 kg m s-1.
Ans (b).
Speed of the electron, v = 6.496 x 106 m/s
Ans (c).
De Broglie wavelength of an electron having a momentum p, is given as:
λ = h/p = ( 6.6 x 10⁻34) /( 5.91 x 10-24) = 1.116 x 10⁻¹⁰m
= 0.112 nm
Therefore, the de Broglie wavelength of the electron is 0.112 nm.
See lessCalculate the (a) momentum, and (b) de Broglie wavelength of the electrons accelerated through a potential difference of 56 V.
Potential difference, V = 56 V Planck's constant, h = 6.6 x 10⁻34 Js Mass of an electron, m = 9.1 x 10⁻31 kg Charge on an electron, e = 1.6 x 10⁻19 C Ans (a). At equilibrium, the kinetic energy of each electron is equal to the accelerating potential, i.e., we can write the relation for velocity (v)Read more
Potential difference, V = 56 V
Planck’s constant, h = 6.6 x 10⁻34 Js
Mass of an electron, m = 9.1 x 10⁻31 kg
Charge on an electron, e = 1.6 x 10⁻19 C
Ans (a).
At equilibrium, the kinetic energy of each electron is equal to the accelerating potential, i.e., we can write the relation for velocity (v) of each electron as:
-1/2mv2=eV
v2 =2eV/m
Therefore ,v = √ (2 x 1.6 x10⁻19 x 56)/ (9.1 x 10⁻31)
= √(19.69 x 10¹² )=4.44 x 10⁶ m/s
The momentum of each accelerated electron is given as:
p = mv = 9.1 x 10-31 x 4.44 x 106 = 4.04 x 10-24 kg m s⁻1 Therefore, the momentum of each electron is 4.04 x 10-24 kg m s-1.
Ans (b).
De Broglie wavelength of an electron accelerating through a potential V, is given by the relation:
λ = 12.27 Aº/√V
= 12.27 x 10⁻¹⁰ /√56
= 0.1639 nm
Therefore, the de Broglie wavelength of each electron is 0.1639 nm.
See lessLight of wavelength 488 nm is produced by an argon laser which is used in the photoelectric effect. When light from this spectral line is incident on the emitter, the stopping (cut-off) potential of photoelectrons is 0.38 V. Find the work function of the material from which the emitter is made.
Wavelength of light produced by the argon laser, λ=488nm = 488 x 10-9m Stopping potential of the photoelectrons, Vo = 0.38 V We know that 1 eV = 1.6 x 10-19 J, therefore Vo =0.38/1.6 x 10-19 eV Planck’s constant, h = 6.6 x 10-34 Js Charge on an electron, e = 1.6 x 10-19 C Speed of light, c = 3 x 10⁸Read more
Wavelength of light produced by the argon laser,
λ=488nm = 488 x 10-9m
Stopping potential of the photoelectrons,
Vo = 0.38 V
We know that 1 eV = 1.6 x 10-19 J, therefore
Vo =0.38/1.6 x 10-19 eV
Planck’s constant, h = 6.6 x 10-34 Js
Charge on an electron, e = 1.6 x 10-19 C
Speed of light, c = 3 x 10⁸ m/s
From Einstein’s photoelectric effect, we have the relation involving the work function φo of the material of the emitter as:
eVo = hc/λ – φo
Therefore , φo = hc/λ -eVo
= (6.6 x 10-34) x (3 x 10⁸ ) /(1.6 x 10-19) ( 488 x 10-9) – (1.6 x 10-19) x 0.38/(1.6 x 10-19)
= 2.54 -0.38 = 2.16 eV
Therefore, the material with which the emitter is made has the work function of 2.16 eV.
See lessLight of frequency 7.21 × 10¹⁴ Hz is incident on a metal surface. Electrons with a maximum speed of 6.0 × 10⁵ m/s are ejected from the surface. What is the threshold frequency for photoemission of electrons?
Frequency of the incident photon, v =488 nm = 488 x 10-9 m Maximum speed of the electrons, v = 6.0 x 105 m/s Planck’s constant, h = 6.626 x 10-34 Js Mass of an electron, m = 9.1 x 10-31 kg For threshold frequency vo, the relation for kinetic energy is written as: 1/2 (mv2 )=h(v-v0 ) v0 = v -mv2 /2hRead more
Frequency of the incident photon, v =488 nm = 488 x 10-9 m
Maximum speed of the electrons, v = 6.0 x 105 m/s
Planck’s constant, h = 6.626 x 10-34 Js
Mass of an electron, m = 9.1 x 10-31 kg
For threshold frequency vo, the relation for kinetic energy is written as:
1/2 (mv2 )=h(v-v0 )
v0 = v -mv2 /2h
= (7.21 x 1014 ) – (9.1 x 10-31) x (6.0 x 105)2 /2 x ( 6.626 x 10-34 )
= (7.21 x 1014 ) -2.472 x 1014
= 4.738 x 1014 Hz
Therefore, the threshold frequency for the photoemission of electrons is 4.738 x 1014 Hz.
See lessThe work function for a certain metal is 4.2 eV. Will this metal give photoelectric emission for incident radiation of wavelength 330 nm?
No Work function of the metal, φO = 4.2 eV Charge on an electron, e = 1.6 x 10-19 C Planck’s constant, h = 6.626 x 10-34 Js Wavelength of the incident radiation, λ = 330 nm = 330 x 10⁻9 m Speed of light, c = 3 x 108 m/s The energy of the incident photon is given as: E = hc/λ = (6.626 x 10-34) x ( 3Read more
No
Work function of the metal, φO = 4.2 eV
Charge on an electron, e = 1.6 x 10-19 C
Planck’s constant, h = 6.626 x 10–34 Js
Wavelength of the incident radiation, λ = 330 nm = 330 x 10⁻9 m
Speed of light, c = 3 x 108 m/s
The energy of the incident photon is given as:
E = hc/λ = (6.626 x 10–34) x ( 3 x 108) /(330 x 10⁻9)
=6.0 x 10⁻19J
=(6.0 x 10⁻19J)/(1.6 x 10⁻19 ) = 3.76 eV
It can be observed that the energy of the incident radiation is less than the work function of the metal. Hence, no photoelectric emission will take place.
See less