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.
Threshold frequency of the metal, v0 = 3.3 x 1014 Hz Frequency of light incident on the metal, v = 8.2 x 1014 Hz Charge on an electron, e = 1.6 x 10-19 C Planck's constant, h = 6.626 x 10-34 Js Cut-off voltage for the photoelectric emission from the metal = V0 The equation for the cut-off energy isRead more
Threshold frequency of the metal, v0 = 3.3 x 1014 Hz
Frequency of light incident on the metal, v = 8.2 x 1014 Hz
Charge on an electron, e = 1.6 x 10-19 C
Planck’s constant, h = 6.626 x 10-34 Js
Cut-off voltage for the photoelectric emission from the metal = V0
The equation for the cut-off energy is given as:
eV0=h(v-v0)
V0=h(v-v0)/e = (6.626 x 10-34) x [(8.2 x 1014) -(3.3 x 1014)]/(1.6 x 10-19) = 2.0292 V
Therefore, the cut-off voltage for the photoelectric emission is 2.0292 V.
Power of the sodium lamp, P = 100 W Wavelength of the emitted sodium light, λ = 589 nm = 589 x 10-9 m Planck's constant, h = 6.626 x 10-34Js Speed of light, c = 3 x 108 m/s Ans (a). The energy per photon associated with the sodium light is given as: E = hc/λ = (6.626 x 10-34) x (3 x 108) /(589 x 10-Read more
Power of the sodium lamp, P = 100 W
Wavelength of the emitted sodium light, λ = 589 nm = 589 x 10-9 m
Planck’s constant, h = 6.626 x 10–34Js
Speed of light, c = 3 x 108 m/s
Ans (a).
The energy per photon associated with the sodium light is given as:
E = hc/λ = (6.626 x 10–34) x (3 x 108) /(589 x 10-9) = 3.37 x 10-¹⁹J
= (3.37 x 10-¹⁹)/(1.6 x 10⁻¹⁹ )
=2.11 eV
Ans (b).
Number of photons delivered to the sphere = n.
The equation for power can be written as: P = nE
=> n = P/E= (100)/ (3.37 x 10-¹⁹) = 2.96 x 10²⁰photons/s
Therefore, every second, 2.96 x 102° photons are delivered to the sphere.
The slope of the cut-off voltage (V) versus frequency (v) of an incident light is given as: v/v = 4.12 x 10-15 Vs V is related to frequency by the equation hv = eV Where, e = Charge on an electron = 1.6 x 10-19 C and h = Planck's constant Therefore, h = e x (v/v)= 1.6 x 10-19 x 4.12 x 10-15 = 6.592Read more
The slope of the cut-off voltage (V) versus frequency (v) of an incident light is given as:
v/v = 4.12 x 10-15 Vs
V is related to frequency by the equation hv = eV
Where, e = Charge on an electron = 1.6 x 10-19 C and
h = Planck’s constant Therefore, h = e x (v/v)= 1.6 x 10-19 x 4.12 x 10–15 = 6.592 x 10–34 Js
Hence ,the value of Plank’s constant is 6.592 x 10–34 Js
Energy flux of sunlight reaching the surface of earth,φ= 1.388 x 103 W/m2 Hence, power of sunlight per square metre, P = 1.388 x 103 W Speed of light, c = 3 x 108 m/s Planck’s constant, h = 6.626 x 10⁻34 Js Average wavelength of photons present in sunlight, λ = 550 nm = 550 x 10⁻9 m Number of photonRead more
Energy flux of sunlight reaching the surface of earth,φ= 1.388 x 103 W/m2
Hence, power of sunlight per square metre, P = 1.388 x 103 W
Speed of light, c = 3 x 108 m/s
Planck’s constant, h = 6.626 x 10⁻34 Js
Average wavelength of photons present in sunlight, λ = 550 nm = 550 x 10⁻9 m
Number of photons per square metre incident on earth per second = n
Hence, the equation for power can be written as:
P = nE
Therefore, n = P/E = P λ /hc = (1.388 x 103) x (550 x 10⁻9) /( 6.626 x 10⁻34 ) (3 x 108 )
= 3.84 x 1021 photons/m²/s
Therefore, every second, 3.84 x 1021 photons are incident per square metre on earth.
Wavelength of the monochromatic light, λ = 632.8 nm = 632.8 x 10-9 m Power emitted by the laser, P = 9.42 mW = 9.42 x 10-3 W Planck's constant, h = 6.626 x 10⁻34 Js Speed of light, c = 3 x 10⁸ m/s Mass of a hydrogen atom, m = 1.66 x 10-27 kg Ans (a). The energy of each photon is given as: E = hc/λ =Read more
Wavelength of the monochromatic light, λ = 632.8 nm = 632.8 x 10-9 m
Power emitted by the laser, P = 9.42 mW = 9.42 x 10-3 W
Planck’s constant, h = 6.626 x 10⁻34 Js
Speed of light, c = 3 x 10⁸ m/s
Mass of a hydrogen atom, m = 1.66 x 10-27 kg
Ans (a).
The energy of each photon is given as:
E = hc/λ
= (6.626 x 10⁻34) x ( 3 x 10⁸) /(632.8 x 10-9) = 3.141 x 10-¹⁹
The momentum of each photon is given as:
P = h/λ = (6.626 x 10⁻34)/(632.8 x 10-9) = 1.047 x 10-27kg ms-¹
Ans (b).
Number of photons arriving per second, at a target irradiated by the beam = n
Assume that the beam has a uniform cross-section that is less than the target area.
Hence, the equation for power can be written as:
P = nE
Therefore , n = P/E = (9.42 x 10-3) /( 3.141 x 10-¹⁹) ≈ 3 x 10¹⁶ photon/s
Ans (c ).
Momentum of the hydrogen atom is the same as the momentum of the photon,
p = 1.047 x 10-27kg ms⁻¹
Momentum is given as: p = mv
Where,
v = Speed of the hydrogen atom
Therefore, v = p/m = (1.047 x 10-27) /(1.66 x 10-27) = 0.621 m/s
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 lessThe threshold frequency for a certain metal is 3.3 × 10¹⁴ Hz. If light of frequency 8.2 × 10¹⁴ Hz is incident on the metal, predict the cutoff voltage for the photoelectric emission
Threshold frequency of the metal, v0 = 3.3 x 1014 Hz Frequency of light incident on the metal, v = 8.2 x 1014 Hz Charge on an electron, e = 1.6 x 10-19 C Planck's constant, h = 6.626 x 10-34 Js Cut-off voltage for the photoelectric emission from the metal = V0 The equation for the cut-off energy isRead more
Threshold frequency of the metal, v0 = 3.3 x 1014 Hz
Frequency of light incident on the metal, v = 8.2 x 1014 Hz
Charge on an electron, e = 1.6 x 10-19 C
Planck’s constant, h = 6.626 x 10-34 Js
Cut-off voltage for the photoelectric emission from the metal = V0
The equation for the cut-off energy is given as:
eV0=h(v-v0)
V0=h(v-v0)/e = (6.626 x 10-34) x [(8.2 x 1014) -(3.3 x 1014)]/(1.6 x 10-19) = 2.0292 V
Therefore, the cut-off voltage for the photoelectric emission is 2.0292 V.
See lessA 100W sodium lamp radiates energy uniformly in all directions. The lamp is located at the centre of a large sphere that absorbs all the sodium light which is incident on it. The wavelength of the sodium light is 589 nm. (a) What is the energy per photon associated with the sodium light? (b) At what rate are the photons delivered to the sphere?
Power of the sodium lamp, P = 100 W Wavelength of the emitted sodium light, λ = 589 nm = 589 x 10-9 m Planck's constant, h = 6.626 x 10-34Js Speed of light, c = 3 x 108 m/s Ans (a). The energy per photon associated with the sodium light is given as: E = hc/λ = (6.626 x 10-34) x (3 x 108) /(589 x 10-Read more
Power of the sodium lamp, P = 100 W
Wavelength of the emitted sodium light, λ = 589 nm = 589 x 10-9 m
Planck’s constant, h = 6.626 x 10–34Js
Speed of light, c = 3 x 108 m/s
Ans (a).
The energy per photon associated with the sodium light is given as:
E = hc/λ = (6.626 x 10–34) x (3 x 108) /(589 x 10-9) = 3.37 x 10-¹⁹J
= (3.37 x 10-¹⁹)/(1.6 x 10⁻¹⁹ )
=2.11 eV
Ans (b).
Number of photons delivered to the sphere = n.
The equation for power can be written as: P = nE
=> n = P/E= (100)/ (3.37 x 10-¹⁹) = 2.96 x 10²⁰photons/s
Therefore, every second, 2.96 x 102° photons are delivered to the sphere.
See lessIn an experiment on photoelectric effect, the slope of the cut-off voltage versus frequency of incident light is found to be 4.12 × 10⁻¹⁵ V s. Calculate the value of Planck’s constant
The slope of the cut-off voltage (V) versus frequency (v) of an incident light is given as: v/v = 4.12 x 10-15 Vs V is related to frequency by the equation hv = eV Where, e = Charge on an electron = 1.6 x 10-19 C and h = Planck's constant Therefore, h = e x (v/v)= 1.6 x 10-19 x 4.12 x 10-15 = 6.592Read more
The slope of the cut-off voltage (V) versus frequency (v) of an incident light is given as:
v/v = 4.12 x 10-15 Vs
V is related to frequency by the equation hv = eV
Where, e = Charge on an electron = 1.6 x 10-19 C and
h = Planck’s constant Therefore, h = e x (v/v)= 1.6 x 10-19 x 4.12 x 10–15 = 6.592 x 10–34 Js
Hence ,the value of Plank’s constant is 6.592 x 10–34 Js
See lessThe energy flux of sunlight reaching the surface of the earth is 1.388 × 10³ W/m² . How many photons (nearly) per square metre are incident on the Earth per second? Assume that the photons in the sunlight have an average wavelength of 550 nm
Energy flux of sunlight reaching the surface of earth,φ= 1.388 x 103 W/m2 Hence, power of sunlight per square metre, P = 1.388 x 103 W Speed of light, c = 3 x 108 m/s Planck’s constant, h = 6.626 x 10⁻34 Js Average wavelength of photons present in sunlight, λ = 550 nm = 550 x 10⁻9 m Number of photonRead more
Energy flux of sunlight reaching the surface of earth,φ= 1.388 x 103 W/m2
Hence, power of sunlight per square metre, P = 1.388 x 103 W
Speed of light, c = 3 x 108 m/s
Planck’s constant, h = 6.626 x 10⁻34 Js
Average wavelength of photons present in sunlight, λ = 550 nm = 550 x 10⁻9 m
Number of photons per square metre incident on earth per second = n
Hence, the equation for power can be written as:
P = nE
Therefore, n = P/E = P λ /hc = (1.388 x 103) x (550 x 10⁻9) /( 6.626 x 10⁻34 ) (3 x 108 )
= 3.84 x 1021 photons/m²/s
Therefore, every second, 3.84 x 1021 photons are incident per square metre on earth.
See lessMonochromatic light of wavelength 632.8 nm is produced by a helium-neon laser. The power emitted is 9.42 mW. (a) Find the energy and momentum of each photon in the light beam, (b) How many photons per second, on the average, arrive at a target irradiated by this beam? (Assume the beam to have uniform cross-section which is less than the target area), and (c) How fast does a hydrogen atom have to travel in order to have the same momentum as that of the photon?
Wavelength of the monochromatic light, λ = 632.8 nm = 632.8 x 10-9 m Power emitted by the laser, P = 9.42 mW = 9.42 x 10-3 W Planck's constant, h = 6.626 x 10⁻34 Js Speed of light, c = 3 x 10⁸ m/s Mass of a hydrogen atom, m = 1.66 x 10-27 kg Ans (a). The energy of each photon is given as: E = hc/λ =Read more
Wavelength of the monochromatic light, λ = 632.8 nm = 632.8 x 10-9 m
Power emitted by the laser, P = 9.42 mW = 9.42 x 10-3 W
Planck’s constant, h = 6.626 x 10⁻34 Js
Speed of light, c = 3 x 10⁸ m/s
Mass of a hydrogen atom, m = 1.66 x 10-27 kg
Ans (a).
The energy of each photon is given as:
E = hc/λ
= (6.626 x 10⁻34) x ( 3 x 10⁸) /(632.8 x 10-9) = 3.141 x 10-¹⁹
The momentum of each photon is given as:
P = h/λ = (6.626 x 10⁻34)/(632.8 x 10-9) = 1.047 x 10-27kg ms-¹
Ans (b).
Number of photons arriving per second, at a target irradiated by the beam = n
Assume that the beam has a uniform cross-section that is less than the target area.
Hence, the equation for power can be written as:
P = nE
Therefore , n = P/E = (9.42 x 10-3) /( 3.141 x 10-¹⁹) ≈ 3 x 10¹⁶ photon/s
Ans (c ).
Momentum of the hydrogen atom is the same as the momentum of the photon,
p = 1.047 x 10-27kg ms⁻¹
Momentum is given as: p = mv
Where,
v = Speed of the hydrogen atom
Therefore, v = p/m = (1.047 x 10-27) /(1.66 x 10-27) = 0.621 m/s
See less