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A metal plate is placed 2 m acoay from a mono - chromatic light source of 1 mW power. Assuming that an electron in metal collects its energy from a circular area of the plate as larger as 10 atomic diameters in radius, calculate haw long (min) it will take for such a target to ‘soak off, 5 eV of energy for its emission from the metal?
Correct answer is '231.33'. Can you explain this answer?
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Verified Answer
A metal plate is placed 2 m acoay from a mono - chromatic light source...
The intensity of light at a distance of 2m from the source
and as the area of target π x 10-28m2 the energy absorbed by the electron per sec.
So, the time taken by electron to absorb 5 eV energy
and as the area of target π x 10-28m2 the energy absorbed by the electron per sec.
So, the time taken by electron to absorb 5 eV energy
Most Upvoted Answer
A metal plate is placed 2 m acoay from a mono - chromatic light source...
To calculate the time it will take for the electron to collect energy from the circular area of the plate, we need to determine the power incident on the circular area.
The power incident on the circular area can be found using the formula:
Power = Intensity * Area
The intensity of the light can be calculated using the formula:
Intensity = Power / (4 * π * distance^2)
Given:
Power = 1 mW = 0.001 W
Distance = 2 m
Intensity = 0.001 / (4 * π * 2^2) = 0.001 / (4 * 3.1416 * 4) = 0.001 / 50.2656 = 0.00001986 W/m^2
Now, we can calculate the power incident on the circular area:
Power = Intensity * Area
Given:
Radius of circular area = 10 atomic diameters = 10 * 2 * 10^-10 m (Assuming atomic diameter is 2 * 10^-10 m)
Area = π * (radius)^2
= 3.1416 * (10 * 2 * 10^-10)^2
= 3.1416 * (4 * 10^-9)
= 12.5664 * 10^-9
= 1.25664 * 10^-8 m^2
Power = 0.00001986 * 1.25664 * 10^-8
= 2.4999 * 10^-13 W
Now, we can calculate the time it will take for the electron to collect this energy.
Energy = Power * Time
Given:
Power = 2.4999 * 10^-13 W
Rearranging the formula:
Time = Energy / Power
Energy = 1 Joule (as 1 watt-second is equal to 1 Joule)
Time = 1 / (2.4999 * 10^-13)
= 4 * 10^12 seconds
Converting seconds to minutes:
Time in minutes = (4 * 10^12) / 60
= 6.6667 * 10^10 minutes
Therefore, it will take approximately 6.6667 * 10^10 minutes for the electron to collect energy from the circular area of the plate.
The power incident on the circular area can be found using the formula:
Power = Intensity * Area
The intensity of the light can be calculated using the formula:
Intensity = Power / (4 * π * distance^2)
Given:
Power = 1 mW = 0.001 W
Distance = 2 m
Intensity = 0.001 / (4 * π * 2^2) = 0.001 / (4 * 3.1416 * 4) = 0.001 / 50.2656 = 0.00001986 W/m^2
Now, we can calculate the power incident on the circular area:
Power = Intensity * Area
Given:
Radius of circular area = 10 atomic diameters = 10 * 2 * 10^-10 m (Assuming atomic diameter is 2 * 10^-10 m)
Area = π * (radius)^2
= 3.1416 * (10 * 2 * 10^-10)^2
= 3.1416 * (4 * 10^-9)
= 12.5664 * 10^-9
= 1.25664 * 10^-8 m^2
Power = 0.00001986 * 1.25664 * 10^-8
= 2.4999 * 10^-13 W
Now, we can calculate the time it will take for the electron to collect this energy.
Energy = Power * Time
Given:
Power = 2.4999 * 10^-13 W
Rearranging the formula:
Time = Energy / Power
Energy = 1 Joule (as 1 watt-second is equal to 1 Joule)
Time = 1 / (2.4999 * 10^-13)
= 4 * 10^12 seconds
Converting seconds to minutes:
Time in minutes = (4 * 10^12) / 60
= 6.6667 * 10^10 minutes
Therefore, it will take approximately 6.6667 * 10^10 minutes for the electron to collect energy from the circular area of the plate.
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A metal plate is placed 2 m acoay from a mono - chromatic light source...
The intensity of light at a distance of 2m from the source
and as the area of target π x 10-28m2 the energy absorbed by the electron per sec.
So, the time taken by electron to absorb 5 eV energy
and as the area of target π x 10-28m2 the energy absorbed by the electron per sec.
So, the time taken by electron to absorb 5 eV energy
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A metal plate is placed 2 m acoay from a mono - chromatic light source of 1 mW power. Assuming that an electron in metal collects its energy from a circular area of the plate as larger as 10 atomic diameters in radius, calculate haw long (min) it will take for such a target to ‘soak off, 5 eV of energy for its emission from the metal?Correct answer is '231.33'. Can you explain this answer?
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A metal plate is placed 2 m acoay from a mono - chromatic light source of 1 mW power. Assuming that an electron in metal collects its energy from a circular area of the plate as larger as 10 atomic diameters in radius, calculate haw long (min) it will take for such a target to ‘soak off, 5 eV of energy for its emission from the metal?Correct answer is '231.33'. Can you explain this answer? for Physics 2024 is part of Physics preparation. The Question and answers have been prepared according to the Physics exam syllabus. Information about A metal plate is placed 2 m acoay from a mono - chromatic light source of 1 mW power. Assuming that an electron in metal collects its energy from a circular area of the plate as larger as 10 atomic diameters in radius, calculate haw long (min) it will take for such a target to ‘soak off, 5 eV of energy for its emission from the metal?Correct answer is '231.33'. Can you explain this answer? covers all topics & solutions for Physics 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A metal plate is placed 2 m acoay from a mono - chromatic light source of 1 mW power. Assuming that an electron in metal collects its energy from a circular area of the plate as larger as 10 atomic diameters in radius, calculate haw long (min) it will take for such a target to ‘soak off, 5 eV of energy for its emission from the metal?Correct answer is '231.33'. Can you explain this answer?.
A metal plate is placed 2 m acoay from a mono - chromatic light source of 1 mW power. Assuming that an electron in metal collects its energy from a circular area of the plate as larger as 10 atomic diameters in radius, calculate haw long (min) it will take for such a target to ‘soak off, 5 eV of energy for its emission from the metal?Correct answer is '231.33'. Can you explain this answer? for Physics 2024 is part of Physics preparation. The Question and answers have been prepared according to the Physics exam syllabus. Information about A metal plate is placed 2 m acoay from a mono - chromatic light source of 1 mW power. Assuming that an electron in metal collects its energy from a circular area of the plate as larger as 10 atomic diameters in radius, calculate haw long (min) it will take for such a target to ‘soak off, 5 eV of energy for its emission from the metal?Correct answer is '231.33'. Can you explain this answer? covers all topics & solutions for Physics 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A metal plate is placed 2 m acoay from a mono - chromatic light source of 1 mW power. Assuming that an electron in metal collects its energy from a circular area of the plate as larger as 10 atomic diameters in radius, calculate haw long (min) it will take for such a target to ‘soak off, 5 eV of energy for its emission from the metal?Correct answer is '231.33'. Can you explain this answer?.
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