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If f(v) is the Maxwell Boltzman distribution for speed v and v_(p) is the most probable speed then f(v_(p)) is (a) 4e((m)/(2 pi k_(B)T))^(1/2) (b) (e)/(4)((m)/((2 pi kg_(B)T))^(1/2) (c) (e)/(e)((m)/(e)(m)/((2 pi kgT))^(1/2)) (d) (1)/(4e)((m)/((2 pi k_(B)T))^(1/2)?
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If f(v) is the Maxwell Boltzman distribution for speed v and v_(p) is ...
Explanation:
The Maxwell-Boltzmann distribution function for the speeds of particles in a gas is given by:
f(v) = 4π( (m)/(2πkT) )^(3/2) v^2 e^((-m*v^2)/(2kT))
where m is the mass of the particles, k is the Boltzmann constant, T is the temperature, and v is the speed of the particles.
The most probable speed v_p is the speed at which the distribution function has its maximum value. This occurs when the derivative of the distribution function with respect to v is zero:
df/dv = 8π( (m)/(2πkT) )^(3/2) v e^((-m*v^2)/(2kT)) - 4π( (m)/(2πkT) )^(3/2) v^3 e^((-m*v^2)/(2kT)) = 0
Simplifying this equation gives:
v_p = (2kT/m)^(1/2)
Solution:
Substituting this value of v_p into the Maxwell-Boltzmann distribution function gives:
f(v_p) = 4π( (m)/(2πkT) )^(3/2) (2kT/m) e^((-m*(2kT/m))/(2kT))
f(v_p) = 4π( (m)/(2πkT) )^(1/2) e^(-1)
f(v_p) = (4/e)( (m)/(2πkT) )^(1/2)
Therefore, the correct option is (a) 4e((m)/(2 pi k_(B)T))^(1/2).
The Maxwell-Boltzmann distribution function for the speeds of particles in a gas is given by:
f(v) = 4π( (m)/(2πkT) )^(3/2) v^2 e^((-m*v^2)/(2kT))
where m is the mass of the particles, k is the Boltzmann constant, T is the temperature, and v is the speed of the particles.
The most probable speed v_p is the speed at which the distribution function has its maximum value. This occurs when the derivative of the distribution function with respect to v is zero:
df/dv = 8π( (m)/(2πkT) )^(3/2) v e^((-m*v^2)/(2kT)) - 4π( (m)/(2πkT) )^(3/2) v^3 e^((-m*v^2)/(2kT)) = 0
Simplifying this equation gives:
v_p = (2kT/m)^(1/2)
Solution:
Substituting this value of v_p into the Maxwell-Boltzmann distribution function gives:
f(v_p) = 4π( (m)/(2πkT) )^(3/2) (2kT/m) e^((-m*(2kT/m))/(2kT))
f(v_p) = 4π( (m)/(2πkT) )^(1/2) e^(-1)
f(v_p) = (4/e)( (m)/(2πkT) )^(1/2)
Therefore, the correct option is (a) 4e((m)/(2 pi k_(B)T))^(1/2).
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If f(v) is the Maxwell Boltzman distribution for speed v and v_(p) is the most probable speed then f(v_(p)) is (a) 4e((m)/(2 pi k_(B)T))^(1/2) (b) (e)/(4)((m)/((2 pi kg_(B)T))^(1/2) (c) (e)/(e)((m)/(e)(m)/((2 pi kgT))^(1/2)) (d) (1)/(4e)((m)/((2 pi k_(B)T))^(1/2)?
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If f(v) is the Maxwell Boltzman distribution for speed v and v_(p) is the most probable speed then f(v_(p)) is (a) 4e((m)/(2 pi k_(B)T))^(1/2) (b) (e)/(4)((m)/((2 pi kg_(B)T))^(1/2) (c) (e)/(e)((m)/(e)(m)/((2 pi kgT))^(1/2)) (d) (1)/(4e)((m)/((2 pi k_(B)T))^(1/2)? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about If f(v) is the Maxwell Boltzman distribution for speed v and v_(p) is the most probable speed then f(v_(p)) is (a) 4e((m)/(2 pi k_(B)T))^(1/2) (b) (e)/(4)((m)/((2 pi kg_(B)T))^(1/2) (c) (e)/(e)((m)/(e)(m)/((2 pi kgT))^(1/2)) (d) (1)/(4e)((m)/((2 pi k_(B)T))^(1/2)? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for If f(v) is the Maxwell Boltzman distribution for speed v and v_(p) is the most probable speed then f(v_(p)) is (a) 4e((m)/(2 pi k_(B)T))^(1/2) (b) (e)/(4)((m)/((2 pi kg_(B)T))^(1/2) (c) (e)/(e)((m)/(e)(m)/((2 pi kgT))^(1/2)) (d) (1)/(4e)((m)/((2 pi k_(B)T))^(1/2)?.
If f(v) is the Maxwell Boltzman distribution for speed v and v_(p) is the most probable speed then f(v_(p)) is (a) 4e((m)/(2 pi k_(B)T))^(1/2) (b) (e)/(4)((m)/((2 pi kg_(B)T))^(1/2) (c) (e)/(e)((m)/(e)(m)/((2 pi kgT))^(1/2)) (d) (1)/(4e)((m)/((2 pi k_(B)T))^(1/2)? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about If f(v) is the Maxwell Boltzman distribution for speed v and v_(p) is the most probable speed then f(v_(p)) is (a) 4e((m)/(2 pi k_(B)T))^(1/2) (b) (e)/(4)((m)/((2 pi kg_(B)T))^(1/2) (c) (e)/(e)((m)/(e)(m)/((2 pi kgT))^(1/2)) (d) (1)/(4e)((m)/((2 pi k_(B)T))^(1/2)? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for If f(v) is the Maxwell Boltzman distribution for speed v and v_(p) is the most probable speed then f(v_(p)) is (a) 4e((m)/(2 pi k_(B)T))^(1/2) (b) (e)/(4)((m)/((2 pi kg_(B)T))^(1/2) (c) (e)/(e)((m)/(e)(m)/((2 pi kgT))^(1/2)) (d) (1)/(4e)((m)/((2 pi k_(B)T))^(1/2)?.
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