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Determine the angular rotation velocity of an S_(2) molecule promoted to the first exccited rotational level if the distance between its nuclei is d= 189p m |
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Answer» Solution :In the first EXCITED raotaional level `J=1` so `E_(J)=1xx2( ħ^(2))/(2I)=(1)/(2)I omega^(2)` classically Thus `omega= sqrt(2)( ħ)/(I)` Now `I=Sigma m_(I)r_(i)^(2)=(m)/(2)(d^(2))/(4)+(m)/(2)(d^(2))/(4)=m(d^(2))/(4)` where `m` is the mass of the mole CUB and `r_(i)` is the distance of the atom from the AXIS. Thus `omega=(4sqrt(2) ħ)/(md^(2))= 1.56xx10^(11) rad//s` |
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