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The key feature of Bohr's spectrum of hy...

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.
A diatomic molecule has moment of inertia `I`. By Bohr's quantization condition its rotational energy in the `n^(th)` level (`n = 0` is not allowed ) is

A

`(1)/(n^(2))((h^(2))/(8pi^(2)I))`

B

`(1)/(n)((h^(2))/(8pi^(2)I))`

C

`n((h^(2))/(8pi^(2)I))`

D

`n^(2)((h^(2))/(8pi^(2)I))`

Text Solution

Verified by Experts

The correct Answer is:
D
`E_(n)=1/2 I omega^(2)=((I omega)^(2))/(2 I)=((nh//2pi)^(2))/(2 I)=(n^(2)h^(2))/(8pi^(2) I)`
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Knowledge Check

  • 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 quntized rotantized rotational energy of a diatomic molecule assuming it to be right . The rate to energy applied is Bohr's quantization condition A diatomic molecute has moment of inertie 1 by Bohr's quantization condition its rotational energy in the n^(th) level (n = 0 is not allowed ) is

    A
    `(1)/(n^(2)) ((h^(2))/(8 pi ^(2) 1))`
    B
    `(1)/(n) ((h^(2))/(8 pi ^(2) 1))`
    C
    `n ((h^(2))/(8 pi ^(2) 1))`
    D
    `n^(2) ((h^(2))/(8 pi ^(2) 1))`

  • A diatomic molecule has moment of inertia I. By applying Bohr's quantisation condition, its rotational energy in the nth level (n = 0 is not allowed) is

    A
    `(1)/(n^(2))((h^(2))/(8pi^(2)I))`
    B
    `(1)/(n)((h^(2))/(8pi^(2)I))`
    C
    `n((h^(2))/(8pi^(2)I))`
    D
    `n^(2)((h^(2))/(8pi^(2)I))`

  • 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 quntized rotantized rotational energy of a diatomic molecule assuming it to be right . The rate to energy applied is Bohr's quantization condition it is found that the excitation from ground to the first excited state of rotation for the CO molecule is close to (4)/(pi) xx 10^(11) Hz then the moment of inertia of CO molecule about its center of mass is close to (Take h = 2 pi xx 10^(-34) J s )

    A
    `2.76 xx 10^(-46) kg m^(2)`
    B
    `1.87 xx 10^(-46) kg m^(2)`
    C
    `4.67 xx 10^(-47) kg m^(2)`
    D
    `1.17 xx 10^(-47) kg m^(2)`

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    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 quntized rotantized rotational energy of a diatomic molecule assuming it to be right . The rate to energy 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

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