Jmax for a rigid diatomic molecule is 3 at 300 K. Calculate the value of rotational constant (in cm-1, rounded up to first decimal place):Correct answer is between '8.4,8.6'. Can you explain this answer?

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Calculation of Rotational Constant

Jmax value: Given Jmax value for a rigid diatomic molecule is 3 at 300 K.

Rotational energy: Rotational energy of a diatomic molecule is given by the expression: E = J(J+1)h^2/8π^2I.

Value of h and π: h = 6.626 x 10^-34 Js and π = 3.14.

Value of moment of inertia: Moment of inertia (I) of a rigid diatomic molecule is given by the expression: I = µr^2, where µ is the reduced mass of the molecule and r is the internuclear distance. For a rigid diatomic molecule, r is constant and µ = m/2, where m is the mass of each atom. For molecules containing isotopes, the reduced mass has to be calculated as: µ = [(m1m2)/(m1+m2)].

Value of reduced mass: For a rigid diatomic molecule, the reduced mass is given by: µ = m/2. The mass of each atom can be found in the periodic table.

Value of rotational constant: The rotational constant (B) is given by the expression: B = h/8π^2I.

Substituting values: Substituting the values of J, h, π, and I in the expression for rotational energy, we get: E = 3(3+1)(6.626 x 10^-34 Js)^2/8π^2(2µr^2). Simplifying this expression, we get: E = 4.98 x 10^-23 J. Converting this energy to cm^-1, we get: E = 0.033 cm^-1. Substituting the values of h, π, and I in the expression for rotational constant, we get: B = (6.626 x 10^-34 Js)/(8π^2(2µr^2)). Substituting the values of π, µ, and r, we get: B = 8.5 cm^-1.

Rounding up: Rounding up this value to the first decimal place, we get: B = 8.5 cm^-1, which falls between the given range of 8.4 and 8.6 cm^-1.

Therefore, the value of rotational constant (in cm^-1, rounded up to the first decimal place) for a rigid diatomic molecule with Jmax value of 3 at 300 K is 8.5 cm^-1, which falls between the given range of 8.4 and 8.6 cm^-1.

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