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Find the angular velocity of rotation of a hydrogen molecule on the first excited rotational level, if the distance between the centres of its atoms is 0.74 Å. |
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Answer» The KINETIC energy will accordingly be `K_(rot)=L^(2)/(2J)=(l(l+1)h^(2))/(2J)` where J is the moment of INERTIA. The kinetic energy in the first excited state is `L_(1)=hsqrt2` and the angular velocity is `omega_(1)=L_(1)/J=(2hsqrt2)/(md^(2))` Here d = 0.74 Å is the distance between the centres of the atoms in the molecule, and m = `1.67xx10^(-27)Kg` is the mass of HYDROGEN atom. |
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