The key feature of Bohr's spectrum of hydrogen atom is the quantization of angular momentum when an electron is revolving around a proton. We will extend this to a general rotational motion to find quantized rotational energy of a diatomic molecule assuming it to be rigid.The rule to be applied is Bohr's quantization condition.
In a `CO` molecule, the distance between `C (mass = 12 a. m. u ) and O (mass = 16 a.m.u)` where `1 a.m.u = (5)/(3) xx 10^(-27) kg , `is close to

Options

A. `2.4xx10^(-10)m`
B. `1.9xx10^(-10)m`
C. `1.3xx10^(-10)m`
D. `4.4xx10^(-11)m`

Answer

C

Explanation

`m_(1)r_(1) = m_(2)r_(2)`
`12 r_(1) = 16 r_(2)`
`(r_(1))/(r_(2)) = (4)/(3) rArr (r_(1))/(l) = (4)/(7)`
`r_(1) = (4)/(7)l`
Now, `I = m_(1)r_(1)^(2)+m_(2)r_(2)^(2)`
`= m_(1)r_(1)(l)`
`= m_(1)((4)/(7)l)l`
`I = ((4m_(1))/(7))l^(2) rArr l = sqrt((2I)/(4m_(1)))`
`l = sqrt((7 xx 1.87 xx 10^(-46))/(4 xx 12 xx (5)/(3) xx 10^(-27)))`
`= 0.128 xx 10^(-9) m` , `= 1.28 xx 10^(-10) m`