The rotational energy of a rigid diatomic molecule can be given by the expression:

E(J) = J(J+1)h^2/8π^2I

Where,
J is the rotational quantum number
h is Planck's constant
I is the moment of inertia of the molecule.



The moment of inertia of a rigid diatomic molecule can be given by the expression:

I = μr^2

Where,
μ is the reduced mass of the molecule
r is the bond length of the molecule.



The reduced mass of a rigid diatomic molecule can be given by the expression:

μ = m1m2/(m1+m2)

Where,
m1 and m2 are the masses of the two atoms in the molecule.



The maximum value of J for a rigid diatomic molecule can be calculated by equating the rotational energy of the molecule at Jmax to the thermal energy at the temperature T.

Jmax(Jmax+1)h^2/8π^2I = kT

Where,
k is the Boltzmann constant
T is the temperature in Kelvin.

Solving for Jmax, we get:

Jmax = (8π^2IkT/h^2)^0.5 - 0.5



Given, rotational constant B = 1.566 cm^-1 at 300 K

Using the relation B = h/8π^2I, we can calculate moment of inertia I as:

I = h/8π^2B = 6.369 x 10^-47 kg m^2

Using the atomic masses of the two atoms in the molecule, we can calculate the reduced mass μ as:

μ = m1m2/(m1+m2) = (m/2)(m/2)/(m/2+m/2) = m/4

Where m is the mass of each atom.

Substituting the values of I, μ, h, k, and T in the equation for Jmax, we get:

Jmax = (8π^2 x 6.369 x 10^-47 x 1.38 x 10^-23 x 300 / 6.626 x 10^-34)^0.5 - 0.5

Jmax = 15.9

Therefore, the maximum value of J for the given rigid diatomic molecule at 300 K is 15.9.

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