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A stream of oxygen molecules at 500 K exists from a pin-hole in an over and strikes a slit that selects the molecules travelling in a specific direction. Given that the pressure outside the over is 2.5 x 10-7 atm, estimate the maximum distance at which the slit must be placed from the pin-hole in order to produce a collimated beam to oxygen, (Radius of O2 = 1.8 x 10-10 m).?
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A stream of oxygen molecules at 500 K exists from a pin-hole in an ove...
Introduction
To produce a collimated beam of oxygen molecules, the distance from the pin-hole to the slit must be calculated based on the molecular velocity and the pressure conditions.
Key Parameters
- Temperature (T): 500 K
- Pressure (P): 2.5 x 10^-7 atm
- Radius of O2 molecule: 1.8 x 10^-10 m
Molecular Velocity Calculation
- Use the formula for average molecular speed (v):
- v = √(3kT/m), where k is the Boltzmann constant and m is the mass of an O2 molecule.
- Mass of O2:
- Approximately 32 g/mol, converting to kg: m = 32 x 10^-3 kg/mol / (6.022 x 10^23 mol^-1) = 5.31 x 10^-26 kg.
- Boltzmann Constant (k):
- k = 1.38 x 10^-23 J/K.
Calculating Distance (L)
- Mean Free Path (λ):
- λ = kT/(√2 * π * d^2 * P), where d is the diameter of the O2 molecule.
- Diameter = 2 * Radius = 3.6 x 10^-10 m.
- Substituting values:
- λ = (1.38 x 10^-23 * 500)/(√2 * π * (3.6 x 10^-10)^2 * 2.5 x 10^-7).
- Final Distance (L):
- For a collimated beam, L should be approximately equal to λ or a few times λ, typically around 5λ for better collimation.
Conclusion
By calculating the mean free path and ensuring the slit is placed at an optimal distance (around 5λ), we can achieve a collimated beam of oxygen molecules from the pin-hole. This placement ensures that the molecules travel in a parallel direction, minimizing scattering and divergence.
To produce a collimated beam of oxygen molecules, the distance from the pin-hole to the slit must be calculated based on the molecular velocity and the pressure conditions.
Key Parameters
- Temperature (T): 500 K
- Pressure (P): 2.5 x 10^-7 atm
- Radius of O2 molecule: 1.8 x 10^-10 m
Molecular Velocity Calculation
- Use the formula for average molecular speed (v):
- v = √(3kT/m), where k is the Boltzmann constant and m is the mass of an O2 molecule.
- Mass of O2:
- Approximately 32 g/mol, converting to kg: m = 32 x 10^-3 kg/mol / (6.022 x 10^23 mol^-1) = 5.31 x 10^-26 kg.
- Boltzmann Constant (k):
- k = 1.38 x 10^-23 J/K.
Calculating Distance (L)
- Mean Free Path (λ):
- λ = kT/(√2 * π * d^2 * P), where d is the diameter of the O2 molecule.
- Diameter = 2 * Radius = 3.6 x 10^-10 m.
- Substituting values:
- λ = (1.38 x 10^-23 * 500)/(√2 * π * (3.6 x 10^-10)^2 * 2.5 x 10^-7).
- Final Distance (L):
- For a collimated beam, L should be approximately equal to λ or a few times λ, typically around 5λ for better collimation.
Conclusion
By calculating the mean free path and ensuring the slit is placed at an optimal distance (around 5λ), we can achieve a collimated beam of oxygen molecules from the pin-hole. This placement ensures that the molecules travel in a parallel direction, minimizing scattering and divergence.
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A stream of oxygen molecules at 500 K exists from a pin-hole in an over and strikes a slit that selects the molecules travelling in a specific direction. Given that the pressure outside the over is 2.5 x 10-7 atm, estimate the maximum distance at which the slit must be placed from the pin-hole in order to produce a collimated beam to oxygen, (Radius of O2 = 1.8 x 10-10 m).?
Question Description
A stream of oxygen molecules at 500 K exists from a pin-hole in an over and strikes a slit that selects the molecules travelling in a specific direction. Given that the pressure outside the over is 2.5 x 10-7 atm, estimate the maximum distance at which the slit must be placed from the pin-hole in order to produce a collimated beam to oxygen, (Radius of O2 = 1.8 x 10-10 m).? for UPSC 2024 is part of UPSC preparation. The Question and answers have been prepared according to the UPSC exam syllabus. Information about A stream of oxygen molecules at 500 K exists from a pin-hole in an over and strikes a slit that selects the molecules travelling in a specific direction. Given that the pressure outside the over is 2.5 x 10-7 atm, estimate the maximum distance at which the slit must be placed from the pin-hole in order to produce a collimated beam to oxygen, (Radius of O2 = 1.8 x 10-10 m).? covers all topics & solutions for UPSC 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A stream of oxygen molecules at 500 K exists from a pin-hole in an over and strikes a slit that selects the molecules travelling in a specific direction. Given that the pressure outside the over is 2.5 x 10-7 atm, estimate the maximum distance at which the slit must be placed from the pin-hole in order to produce a collimated beam to oxygen, (Radius of O2 = 1.8 x 10-10 m).?.
A stream of oxygen molecules at 500 K exists from a pin-hole in an over and strikes a slit that selects the molecules travelling in a specific direction. Given that the pressure outside the over is 2.5 x 10-7 atm, estimate the maximum distance at which the slit must be placed from the pin-hole in order to produce a collimated beam to oxygen, (Radius of O2 = 1.8 x 10-10 m).? for UPSC 2024 is part of UPSC preparation. The Question and answers have been prepared according to the UPSC exam syllabus. Information about A stream of oxygen molecules at 500 K exists from a pin-hole in an over and strikes a slit that selects the molecules travelling in a specific direction. Given that the pressure outside the over is 2.5 x 10-7 atm, estimate the maximum distance at which the slit must be placed from the pin-hole in order to produce a collimated beam to oxygen, (Radius of O2 = 1.8 x 10-10 m).? covers all topics & solutions for UPSC 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A stream of oxygen molecules at 500 K exists from a pin-hole in an over and strikes a slit that selects the molecules travelling in a specific direction. Given that the pressure outside the over is 2.5 x 10-7 atm, estimate the maximum distance at which the slit must be placed from the pin-hole in order to produce a collimated beam to oxygen, (Radius of O2 = 1.8 x 10-10 m).?.
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A stream of oxygen molecules at 500 K exists from a pin-hole in an over and strikes a slit that selects the molecules travelling in a specific direction. Given that the pressure outside the over is 2.5 x 10-7 atm, estimate the maximum distance at which the slit must be placed from the pin-hole in order to produce a collimated beam to oxygen, (Radius of O2 = 1.8 x 10-10 m).?, a detailed solution for A stream of oxygen molecules at 500 K exists from a pin-hole in an over and strikes a slit that selects the molecules travelling in a specific direction. Given that the pressure outside the over is 2.5 x 10-7 atm, estimate the maximum distance at which the slit must be placed from the pin-hole in order to produce a collimated beam to oxygen, (Radius of O2 = 1.8 x 10-10 m).? has been provided alongside types of A stream of oxygen molecules at 500 K exists from a pin-hole in an over and strikes a slit that selects the molecules travelling in a specific direction. Given that the pressure outside the over is 2.5 x 10-7 atm, estimate the maximum distance at which the slit must be placed from the pin-hole in order to produce a collimated beam to oxygen, (Radius of O2 = 1.8 x 10-10 m).? theory, EduRev gives you an
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