When a hypothetical diatomic molecule having atoms 0.8870 nm apart undergoes a rotational transition from the Iota = 2 state to the next lower state, it gives up a photon having energy 8.880 times 10^-4 eV. the force constant of this molecule
k' = 3.188 * 10^-9 N/m.
Generally, The energy of a rotational transition is given by the equation:
E = hcB'J
where h is the Planck constant, c is the speed of light, B' is the rotational constant, and J is the rotational quantum number. Similarly, the energy of a vibrational transition is given by:
E = hcnu
where h is the Planck constant, c is the speed of light, nu is the vibrational frequency, and u is the vibrational quantum number.
We can use these equations to solve for the rotational and vibrational constants for the molecule. First, we'll solve for the rotational constant:
E = hcB'J 8.880 * 10^-4 eV = (6.626 * 10^-34 J * s) * (2.998 * 10^8 m/s) * B' * 2 B' = 1.478 * 10^-40 m^2
Next, we'll solve for the vibrational frequency:
E = hcnu 0.2560 eV = (6.626 * 10^-34 J * s) * (2.998 * 10^8 m/s) * nu nu = 3.238 * 10^13 s^-1
Finally, we can use the vibrational frequency to solve for the force constant:
nu = sqrt(k'/(mu)) k' = (nu^2) * (mu) k' = (3.238 * 10^13 s^-1)^2 * (2 * 0.8870 * 10^-9 kg) k' = 3.188 * 10^-9 N/m
So the force constant of the molecule is k' = 3.188 * 10^-9 N/m.
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