Animated Solution for Physics - Atoms and Nuclei: Comprehension Passage
The key feature of Bohr's theory of spectrum of hydrogen atom is the quantisation of angular momentum when an electron is revolving around a proton. We will extend this to a general rotational motion to find quantised rotational energy of a diatomic molecule assuming it to be rigid. The rule to be applied is Bohr's quantisation condition.
Question 1:
A diatomic molecule has moment of inertia I. By Bohr's quantization condition its rotational energy in the nth level (n=0 is not allowed) is
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Question 2:
It is found that the excitation frequency from ground to the first excited state of rotation for the CO molecule is close to π4×1011 Hz. Then the moment of inertia of CO molecule about its centre of mass is close to (Take h=2π×10−34 J-s)
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Question 3:
In a CO molecule, the distance between C (mass =12 amu) and O (mass =16 amu), where 1 amu=35×10−27 kg, is close to
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Visualized Solution
DiatomicMoleculeRotation
Diatomic molecule rotating about its Center of Mass (COM)
Bohr′sQuantizationCondition
L=Iω
L=2πnh
⇒Iω=2πnh
RotationalKineticEnergy
K=21Iω2
QuantizedEnergyLevels
Kn=21I(2πInh)2
Kn=8π2In2h2
EnergyStates
Ground state: n=1
First excited state: n=2
EnergyDifference
ΔK=K2−K1
ΔK=8π2I22h2−8π2I12h2
ΔK=8π2I3h2
SolvingforMomentofInertia
hf=8π2I3h2
I=8π2f3h
SubstitutingValues
I=8π2(π4×1011)3(2π×10−34)
CalculatingI
I=32π×10116π×10−34
I=1.875×10−46 kg-m2
ReducedMassandBondLength
I=μr2
μ=m1+m2m1m2
CalculatingReducedMass(amu)
μ=12+1612×16 amu
μ=748 amu
ConvertingReducedMasstokg
μ=748×35×10−27 kg
μ=780×10−27 kg≈11.43×10−27 kg
SolvingforBondLengthr
r=μI
r=11.43×10−271.875×10−46
FinalCalculationforr
r=1.64×10−20 m
r≈1.28×10−10 m
TheRigidRotorApproximation
Rigid Rotor Approximation
Centrifugal distortion increases I at high n
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The Sigma Insight: Bohr's Atomic Model and Energy Levels
Solution Diagram
Quantum Mechanics of a Spinning Molecule
The Rigid Rotor
Imagine a diatomic molecule, like Carbon Monoxide (CO), spinning around its center of mass. Just like an electron orbiting a nucleus in the Bohr model, this macroscopic rotation isn't arbitrary. It strictly follows the rules of quantum mechanics. By extending Bohr's quantization condition to a rotating rigid body, we can unlock the secrets of molecular spectra and even measure the distance between atoms!
The Master Equation
Quantizing Rotation
Bohr's quantization condition states that the angular momentum, L, must be an integral multiple of 2πh. For a rotating rigid body, the angular momentum is the product of its moment of inertia, I, and its angular velocity, ω:
L=Iω=2πnh
Classically, the rotational kinetic energy is given by K=21Iω2. Let's substitute ω from our quantization condition into this energy formula:
Kn=21I(2πInh)2=8π2In2h2
This elegant equation reveals that rotational energy is quantized into discrete levels, perfectly answering our first question.
Decoding the Excitation Frequency
When a molecule absorbs a photon, it jumps to a higher rotational energy state. The ground state for rotation is n=1 (since n=0 would mean zero energy and zero angular momentum, which is not allowed in this simplified model). The first excited state is n=2.
The energy of the absorbed photon, hf, must exactly equal the energy difference between these two states:
ΔK=K2−K1=8π2I22h2−8π2I12h2=8π2I3h2
Equating this to hf, we can solve for the moment of inertia, I:
hf=8π2I3h2⟹I=8π2f3h
Substituting the given values f=π4×1011 Hz and h=2π×10−34 J-s:
This brilliant deduction gives us the moment of inertia of the CO molecule!
Measuring the Unseen
Bond Length
Finally, how do we find the distance between the Carbon and Oxygen atoms? For a diatomic molecule, the moment of inertia about the center of mass is simply the reduced mass, μ, times the square of the bond length, r:
I=μr2
The reduced mass μ is the product of the masses divided by their sum:
μ=m1+m2m1m2=12+1612×16=28192=748 amu
Converting this to kilograms using the given factor 1 amu=35×10−27 kg:
μ=748×35×10−27=780×10−27≈11.43×10−27 kg
Now, we just need to isolate r and plug in our values:
r=μI=11.43×10−271.875×10−46=1.64×10−20≈1.28×10−10 m
And there we have it! By simply observing the frequency of light a molecule absorbs, we have deduced the microscopic distance between its atoms. This is the true power and beauty of quantum mechanics.
