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
The 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