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Light rays of wavelengths 6000 Å and of photon intensity 39 x 6 watts/m2 is incident on a metal surface. If only one percent of photons incident on the surface emit photo electrons, then the number of electrons emitted per second per unit area from the surface will be [ Planck constant = 6.64 x 10-34 J − s ; Velocity of light = 3 x 108 m s-1 ]
- a)2 x 1018
- b)10 x 1018
- c)12 x 1017
- d)12 x 1015
Correct answer is option 'C'. Can you explain this answer?
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Light rays of wavelengths 6000 and of photon intensity 39 x6 watts/m2...
S, speed of light = 3 x 108 m/s, charge of an electron = 1.6 x 10-19 C]
First, we need to calculate the energy of a single photon of wavelength 6000 Å using the equation:
E = hc/λ
where h is Planck's constant, c is the speed of light, and λ is the wavelength.
E = (6.64 x 10^-34 J s) x (3 x 10^8 m/s) / (6000 x 10^-10 m)
E = 3.31 x 10^-19 J
Next, we can calculate the number of photons per second per unit area using the intensity:
I = P/A = N x E/t x A
where P is the power, A is the area, N is the number of photons per second, E is the energy per photon, and t is the time. We can rearrange this equation to solve for N:
N = I x A / E
N = (39 x 6 W/m2) x (1/3.31 x 10^-19 J)
N ≈ 7.03 x 10^20 photons/s/m2
Since only 1% of these photons will emit photoelectrons, we can calculate the number of electrons emitted per second per unit area using the photoelectric effect equation:
I = Ne/t
where I is the current, N is the number of electrons emitted per second, e is the charge of an electron, and t is the time. We can rearrange this equation to solve for N:
N = I x t / e
Since we're given the intensity in watts/m2, we can assume that the entire area is emitting electrons, so A = 1 m2. We'll also assume a time t of 1 second. Plugging in the values, we get:
N = (0.01) x (7.03 x 10^20 photons/s/m2) x (1 second) / (1.6 x 10^-19 C)
N ≈ 4.39 x 10^13 electrons/s/m2
Therefore, the number of electrons emitted per second per unit area from the surface is approximately 4.39 x 10^13 electrons/s/m2.
First, we need to calculate the energy of a single photon of wavelength 6000 Å using the equation:
E = hc/λ
where h is Planck's constant, c is the speed of light, and λ is the wavelength.
E = (6.64 x 10^-34 J s) x (3 x 10^8 m/s) / (6000 x 10^-10 m)
E = 3.31 x 10^-19 J
Next, we can calculate the number of photons per second per unit area using the intensity:
I = P/A = N x E/t x A
where P is the power, A is the area, N is the number of photons per second, E is the energy per photon, and t is the time. We can rearrange this equation to solve for N:
N = I x A / E
N = (39 x 6 W/m2) x (1/3.31 x 10^-19 J)
N ≈ 7.03 x 10^20 photons/s/m2
Since only 1% of these photons will emit photoelectrons, we can calculate the number of electrons emitted per second per unit area using the photoelectric effect equation:
I = Ne/t
where I is the current, N is the number of electrons emitted per second, e is the charge of an electron, and t is the time. We can rearrange this equation to solve for N:
N = I x t / e
Since we're given the intensity in watts/m2, we can assume that the entire area is emitting electrons, so A = 1 m2. We'll also assume a time t of 1 second. Plugging in the values, we get:
N = (0.01) x (7.03 x 10^20 photons/s/m2) x (1 second) / (1.6 x 10^-19 C)
N ≈ 4.39 x 10^13 electrons/s/m2
Therefore, the number of electrons emitted per second per unit area from the surface is approximately 4.39 x 10^13 electrons/s/m2.
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Light rays of wavelengths 6000 and of photon intensity 39 x6 watts/m2 is incident on a metal surface. If only one percent of photons incident on the surface emit photo electrons, then the number of electrons emitted per second per unit area from the surface will be [ Planck constant = 6.64 x10-34 J − s ; Velocity of light = 3 x108 m s-1]a)2 x 1018b)10 x 1018c)12 x 1017d)12 x 1015Correct answer is option 'C'. Can you explain this answer?
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Light rays of wavelengths 6000 and of photon intensity 39 x6 watts/m2 is incident on a metal surface. If only one percent of photons incident on the surface emit photo electrons, then the number of electrons emitted per second per unit area from the surface will be [ Planck constant = 6.64 x10-34 J − s ; Velocity of light = 3 x108 m s-1]a)2 x 1018b)10 x 1018c)12 x 1017d)12 x 1015Correct answer is option 'C'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about Light rays of wavelengths 6000 and of photon intensity 39 x6 watts/m2 is incident on a metal surface. If only one percent of photons incident on the surface emit photo electrons, then the number of electrons emitted per second per unit area from the surface will be [ Planck constant = 6.64 x10-34 J − s ; Velocity of light = 3 x108 m s-1]a)2 x 1018b)10 x 1018c)12 x 1017d)12 x 1015Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Light rays of wavelengths 6000 and of photon intensity 39 x6 watts/m2 is incident on a metal surface. If only one percent of photons incident on the surface emit photo electrons, then the number of electrons emitted per second per unit area from the surface will be [ Planck constant = 6.64 x10-34 J − s ; Velocity of light = 3 x108 m s-1]a)2 x 1018b)10 x 1018c)12 x 1017d)12 x 1015Correct answer is option 'C'. Can you explain this answer?.
Light rays of wavelengths 6000 and of photon intensity 39 x6 watts/m2 is incident on a metal surface. If only one percent of photons incident on the surface emit photo electrons, then the number of electrons emitted per second per unit area from the surface will be [ Planck constant = 6.64 x10-34 J − s ; Velocity of light = 3 x108 m s-1]a)2 x 1018b)10 x 1018c)12 x 1017d)12 x 1015Correct answer is option 'C'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about Light rays of wavelengths 6000 and of photon intensity 39 x6 watts/m2 is incident on a metal surface. If only one percent of photons incident on the surface emit photo electrons, then the number of electrons emitted per second per unit area from the surface will be [ Planck constant = 6.64 x10-34 J − s ; Velocity of light = 3 x108 m s-1]a)2 x 1018b)10 x 1018c)12 x 1017d)12 x 1015Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Light rays of wavelengths 6000 and of photon intensity 39 x6 watts/m2 is incident on a metal surface. If only one percent of photons incident on the surface emit photo electrons, then the number of electrons emitted per second per unit area from the surface will be [ Planck constant = 6.64 x10-34 J − s ; Velocity of light = 3 x108 m s-1]a)2 x 1018b)10 x 1018c)12 x 1017d)12 x 1015Correct answer is option 'C'. Can you explain this answer?.
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