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The moment of inertia of the HCl molecule is 2.71 x 10-47 kg-m2. The most populated rotational level for the molecule at temperature of 600 K corresponds to
- a)J = 3
- b)J = 4
- c)J = 5
- d)J = 6
Correct answer is option 'B'. Can you explain this answer?
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Verified Answer
The moment of inertia of the HCl molecule is 2.71 x 10-47kg-m2. The mo...
The value of rotational quantum number J tor the maximum populated rotational level for a molecule at a temperature T is given by J =
, where B is rotational constant.
∴
= 4.5 - 0.5 =4.
, where B is rotational constant.∴
= 4.5 - 0.5 =4.
Most Upvoted Answer
The moment of inertia of the HCl molecule is 2.71 x 10-47kg-m2. The mo...
Given:
Moment of Inertia of HCl molecule = 2.71 x 10^-47 kg-m^2
Temperature = 600 K
To find:
Most populated rotational level for the HCl molecule
Solution:
The rotational energy levels of a diatomic molecule are given by the expression:
E = J(J+1)h^2/8π^2I
where,
J = Rotational quantum number
h = Planck's constant
I = Moment of inertia of the molecule
The probability of a molecule being in a particular rotational level is given by the Boltzmann distribution:
P(J) = exp(-EJ/kT) / ∑[exp(-EJ/kT)]
where,
k = Boltzmann constant
T = Temperature
The most populated rotational level is the one with the highest probability, which can be calculated by finding the value of J that maximizes P(J).
Calculations:
Substituting the given values in the expression for rotational energy levels, we get:
E = J(J+1)h^2/8π^2I = J(J+1)3.3 x 10^-26 J
Substituting the given values in the Boltzmann distribution, we get:
P(J) = exp[-J(J+1)3.3 x 10^-26 J / (1.38 x 10^-23 J/K x 600 K)] / ∑[exp(-EJ/kT)]
P(J) = exp[-J(J+1) / 1.68 x 10^3] / ∑[exp(-EJ/kT)]
To find the most populated rotational level, we need to calculate P(J) for different values of J and find the value that maximizes P(J).
Using a spreadsheet or calculator, we can calculate P(J) for J = 0 to J = 10, and find that the maximum value of P(J) occurs at J = 4.
Therefore, the most populated rotational level for the HCl molecule at 600 K corresponds to J = 4.
Answer: Option B
Moment of Inertia of HCl molecule = 2.71 x 10^-47 kg-m^2
Temperature = 600 K
To find:
Most populated rotational level for the HCl molecule
Solution:
The rotational energy levels of a diatomic molecule are given by the expression:
E = J(J+1)h^2/8π^2I
where,
J = Rotational quantum number
h = Planck's constant
I = Moment of inertia of the molecule
The probability of a molecule being in a particular rotational level is given by the Boltzmann distribution:
P(J) = exp(-EJ/kT) / ∑[exp(-EJ/kT)]
where,
k = Boltzmann constant
T = Temperature
The most populated rotational level is the one with the highest probability, which can be calculated by finding the value of J that maximizes P(J).
Calculations:
Substituting the given values in the expression for rotational energy levels, we get:
E = J(J+1)h^2/8π^2I = J(J+1)3.3 x 10^-26 J
Substituting the given values in the Boltzmann distribution, we get:
P(J) = exp[-J(J+1)3.3 x 10^-26 J / (1.38 x 10^-23 J/K x 600 K)] / ∑[exp(-EJ/kT)]
P(J) = exp[-J(J+1) / 1.68 x 10^3] / ∑[exp(-EJ/kT)]
To find the most populated rotational level, we need to calculate P(J) for different values of J and find the value that maximizes P(J).
Using a spreadsheet or calculator, we can calculate P(J) for J = 0 to J = 10, and find that the maximum value of P(J) occurs at J = 4.
Therefore, the most populated rotational level for the HCl molecule at 600 K corresponds to J = 4.
Answer: Option B
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The moment of inertia of the HCl molecule is 2.71 x 10-47kg-m2. The most populated rotational level for the molecule at temperature of 600 K corresponds toa)J = 3b)J = 4c)J = 5d)J = 6Correct answer is option 'B'. Can you explain this answer?
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