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663 mW of light from a 540 nm source is incident on the surface of a metal. If only 1 of each 5×109 incident photons in absorbed and causes an electron to be ejected from the surface, the total photocurrent in the circuit is ____________________.
- a)5.76 x 1011
- b)5.76 x 109
- c)5.76 x 1010
- d)5.76 x 1012
Correct answer is option 'A'. Can you explain this answer?
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Verified Answer
663 mW of light from a 540 nm source is incident on the surface of a m...
N / Δt = no. of photon incident per second.
∴ 663x10-3=(N/Δt)x(hc/λ)
∴N/ Δt=663x10-3/(hc/λ)= 663x10-3/(1242nmeV/540)
(n/Δt)/(N/Δt)=1/(5x109)
n/Δt=[1/(5x109)]x (N/Δt)
=[1/(5x109)]x ((663x10-3x540)/1242nmeV)
I=ne/Δt=5.76x1011A
∴ 663x10-3=(N/Δt)x(hc/λ)
∴N/ Δt=663x10-3/(hc/λ)= 663x10-3/(1242nmeV/540)
(n/Δt)/(N/Δt)=1/(5x109)
n/Δt=[1/(5x109)]x (N/Δt)
=[1/(5x109)]x ((663x10-3x540)/1242nmeV)
I=ne/Δt=5.76x1011A
Most Upvoted Answer
663 mW of light from a 540 nm source is incident on the surface of a m...
Photons is able to eject an electron from the metal, calculate the number of electrons ejected per second.
To calculate the number of electrons ejected per second, we need to determine the number of photons incident on the metal surface per second and then multiply it by the probability of ejecting an electron.
First, let's calculate the number of photons incident on the metal surface per second.
The power of the light source is given as 663 mW, which is equal to 663 × 10^-3 W. We can use the formula:
Power = Energy / Time
where Energy is the energy of each photon and Time is the time in seconds.
The energy of each photon can be calculated using the equation:
Energy = Planck's constant × frequency
The frequency of the light can be calculated using the equation:
Frequency = speed of light / wavelength
The speed of light is a constant value of approximately 3 × 10^8 m/s.
Let's calculate the frequency and energy of each photon:
Frequency = (3 × 10^8 m/s) / (540 × 10^-9 m)
Frequency = 5.56 × 10^14 Hz
Energy = (6.63 × 10^-34 J·s) × (5.56 × 10^14 Hz)
Energy = 3.68 × 10^-19 J
Now, let's calculate the number of photons incident on the metal surface per second:
Power = Energy / Time
663 × 10^-3 W = (3.68 × 10^-19 J) / Time
Time = (3.68 × 10^-19 J) / (663 × 10^-3 W)
Time = 5.55 × 10^-17 s
The number of photons incident on the metal surface per second can be calculated using the equation:
Number of photons = Power / Energy of each photon
Number of photons = (663 × 10^-3 W) / (3.68 × 10^-19 J)
Number of photons = 1.80 × 10^15 photons
Now, let's calculate the number of electrons ejected per second:
Number of electrons ejected per second = (1/5) × Number of photons
Number of electrons ejected per second = (1/5) × (1.80 × 10^15 photons)
Number of electrons ejected per second = 3.60 × 10^14 electrons
Therefore, approximately 3.60 × 10^14 electrons are ejected per second.
To calculate the number of electrons ejected per second, we need to determine the number of photons incident on the metal surface per second and then multiply it by the probability of ejecting an electron.
First, let's calculate the number of photons incident on the metal surface per second.
The power of the light source is given as 663 mW, which is equal to 663 × 10^-3 W. We can use the formula:
Power = Energy / Time
where Energy is the energy of each photon and Time is the time in seconds.
The energy of each photon can be calculated using the equation:
Energy = Planck's constant × frequency
The frequency of the light can be calculated using the equation:
Frequency = speed of light / wavelength
The speed of light is a constant value of approximately 3 × 10^8 m/s.
Let's calculate the frequency and energy of each photon:
Frequency = (3 × 10^8 m/s) / (540 × 10^-9 m)
Frequency = 5.56 × 10^14 Hz
Energy = (6.63 × 10^-34 J·s) × (5.56 × 10^14 Hz)
Energy = 3.68 × 10^-19 J
Now, let's calculate the number of photons incident on the metal surface per second:
Power = Energy / Time
663 × 10^-3 W = (3.68 × 10^-19 J) / Time
Time = (3.68 × 10^-19 J) / (663 × 10^-3 W)
Time = 5.55 × 10^-17 s
The number of photons incident on the metal surface per second can be calculated using the equation:
Number of photons = Power / Energy of each photon
Number of photons = (663 × 10^-3 W) / (3.68 × 10^-19 J)
Number of photons = 1.80 × 10^15 photons
Now, let's calculate the number of electrons ejected per second:
Number of electrons ejected per second = (1/5) × Number of photons
Number of electrons ejected per second = (1/5) × (1.80 × 10^15 photons)
Number of electrons ejected per second = 3.60 × 10^14 electrons
Therefore, approximately 3.60 × 10^14 electrons are ejected per second.
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663 mW of light from a 540 nm source is incident on the surface of a metal. If only 1 of each 5×109incident photons in absorbed and causes an electron to be ejected from the surface, the total photocurrent in the circuit is ____________________.a)5.76 x 1011b)5.76 x 109c)5.76 x 1010d)5.76 x 1012Correct answer is option 'A'. Can you explain this answer?
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663 mW of light from a 540 nm source is incident on the surface of a metal. If only 1 of each 5×109incident photons in absorbed and causes an electron to be ejected from the surface, the total photocurrent in the circuit is ____________________.a)5.76 x 1011b)5.76 x 109c)5.76 x 1010d)5.76 x 1012Correct answer is option 'A'. Can you explain this answer? for Class 12 2024 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about 663 mW of light from a 540 nm source is incident on the surface of a metal. If only 1 of each 5×109incident photons in absorbed and causes an electron to be ejected from the surface, the total photocurrent in the circuit is ____________________.a)5.76 x 1011b)5.76 x 109c)5.76 x 1010d)5.76 x 1012Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for Class 12 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 663 mW of light from a 540 nm source is incident on the surface of a metal. If only 1 of each 5×109incident photons in absorbed and causes an electron to be ejected from the surface, the total photocurrent in the circuit is ____________________.a)5.76 x 1011b)5.76 x 109c)5.76 x 1010d)5.76 x 1012Correct answer is option 'A'. Can you explain this answer?.
663 mW of light from a 540 nm source is incident on the surface of a metal. If only 1 of each 5×109incident photons in absorbed and causes an electron to be ejected from the surface, the total photocurrent in the circuit is ____________________.a)5.76 x 1011b)5.76 x 109c)5.76 x 1010d)5.76 x 1012Correct answer is option 'A'. Can you explain this answer? for Class 12 2024 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about 663 mW of light from a 540 nm source is incident on the surface of a metal. If only 1 of each 5×109incident photons in absorbed and causes an electron to be ejected from the surface, the total photocurrent in the circuit is ____________________.a)5.76 x 1011b)5.76 x 109c)5.76 x 1010d)5.76 x 1012Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for Class 12 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 663 mW of light from a 540 nm source is incident on the surface of a metal. If only 1 of each 5×109incident photons in absorbed and causes an electron to be ejected from the surface, the total photocurrent in the circuit is ____________________.a)5.76 x 1011b)5.76 x 109c)5.76 x 1010d)5.76 x 1012Correct answer is option 'A'. Can you explain this answer?.
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