It is found that the excitation frequency from ground to the first excited state of rotation for the

Q: Why is the ground state $n=1$ and not $n=0$?

In Bohr's model, $n=0$ would mean zero angular momentum and zero energy, which violates the uncertainty principle. The problem also explicitly states (in the preceding context) that $n=0$ is not allowed.

Q: Why do we use $E_n = \frac{n^2 h^2}{8\pi^2 I}$ instead of the standard quantum mechanical formula $E_J = \frac{J(J+1)h^2}{8\pi^2 I}$?

The problem specifically instructs us to use Bohr's quantization condition ($L = \frac{nh}{2\pi}$), which leads to $E = \frac{L^2}{2I} = \frac{n^2 h^2}{8\pi^2 I}$. We must follow the model specified in the question.

Q: How did we get the energy difference as $\frac{3h^2}{8\pi^2 I}$?

The transition is from $n=1$ to $n=2$. So, $\Delta E = E_2 - E_1 = \frac{2^2 h^2}{8\pi^2 I} - \frac{1^2 h^2}{8\pi^2 I} = \frac{4h^2}{8\pi^2 I} - \frac{h^2}{8\pi^2 I} = \frac{3h^2}{8\pi^2 I}$.

Q: What does the moment of inertia physically represent here?

It represents the rotational inertia of the CO molecule about its center of mass. It depends on the masses of the Carbon and Oxygen atoms and the square of the bond length between them.

Q: Why did the $\pi$ terms cancel out so perfectly?

In many physics problems, especially in JEE, the values are carefully chosen by the examiners to simplify the calculation. Here, $h$ has a $\pi$ in the numerator, and $f$ has a $\pi$ in the denominator, which perfectly cancels the $\pi^2$ in the formula.

Question

It is found that the excitation frequency from ground to the first excited state of rotation for the CO molecule is close to $\frac{4}{\pi} 	imes 10^{11}	ext{ Hz}$. Then the moment of inertia of CO molecule about its centre of mass is close to (Take $h = 2\pi 	imes 10^{-34}	ext{ J-s}$)

Options:

(A) $2.76 	imes 10^{-46}	ext{ kg - m}^2$

(B) $1.87 	imes 10^{-46}	ext{ kg - m}^2$

(C) $4.67 	imes 10^{-47}	ext{ kg - m}^2$

(D) $1.17 	imes 10^{-47}	ext{ kg - m}^2$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine a carbon monoxide (CO) molecule spinning around its center of mass. In classical physics, this molecule could rotate at any speed, possessing any amount of rotational energy. However, the microscopic world plays by different rules.

When we apply **Bohr's quantization condition** to this rotating system, we restrict the angular momentum $L$ to be an integral multiple of $\frac{h}{2\pi}$.

Mathematically, this is written as:
$$L = \frac{nh}{2\pi}$$

Since rotational kinetic energy is given by $E = \frac{L^2}{2I}$, where $I$ is the moment of inertia, substituting our quantized angular momentum gives us the allowed energy levels:
$$E_n = \frac{n^2 h^2}{8\pi^2 I}$$

This is a beautiful result! It tells us that the molecule can only exist in specific, discrete rotational energy states.

### The Master Equation

The problem states that the molecule is excited from the ground state ($n=1$) to the first excited state ($n=2$).

To make this jump, the molecule must absorb a photon whose energy exactly matches the energy difference between these two states.

Let's calculate this energy difference:
$$\Delta E = E_2 - E_1$$
$$\Delta E = \frac{2^2 h^2}{8\pi^2 I} - \frac{1^2 h^2}{8\pi^2 I}$$
$$\Delta E = \frac{3h^2}{8\pi^2 I}$$

We know that the energy of the absorbed photon is $hf$. Equating the two, we get:
$$hf = \frac{3h^2}{8\pi^2 I}$$

We can cancel one $h$ from both sides, leaving us with:
$$f = \frac{3h}{8\pi^2 I}$$

Our goal is to find the moment of inertia, $I$. Rearranging the equation to isolate $I$, we obtain our master equation:
$$I = \frac{3h}{8\pi^2 f}$$

### Final Calculation

Now, it's time for the execution. We are given the values for Planck's constant and the excitation frequency:
$h = 2\pi 	imes 10^{-34} 	ext{ J-s}$
$f = \frac{4}{\pi} 	imes 10^{11} 	ext{ Hz}$

Let's carefully substitute these into our master equation:
$$I = \frac{3(2\pi 	imes 10^{-34})}{8\pi^2 \left(\frac{4}{\pi} 	imes 10^{11}ight)}$$

Notice how elegantly the examiners have set up the numbers! The $\pi$ in the numerator of $h$ and the $\pi$ in the denominator of $f$ will perfectly cancel out the $\pi^2$ in the denominator of our formula.

Let's simplify:
$$I = \frac{6\pi 	imes 10^{-34}}{32\pi 	imes 10^{11}}$$
$$I = \frac{6}{32} 	imes 10^{-45}$$
$$I = \frac{3}{16} 	imes 10^{-45}$$

Dividing $3$ by $16$ gives $0.1875$. 
$$I = 0.1875 	imes 10^{-45} 	ext{ kg-m}^2$$

To match the options, we adjust the decimal point:
$$I = 1.875 	imes 10^{-46} 	ext{ kg-m}^2$$

Rounding to two decimal places, we get **$1.87 	imes 10^{-46} 	ext{ kg-m}^2$**, which perfectly matches option (b). The physics of the microscopic world never fails to amaze!

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Visualized Solution

$Bohr’s Quantization for Rotation$

$Energy of Transition$

$Isolating Moment of Inertia$

$Substituting Values$

$Final Calculation$

$The Way Forward$

The Sigma Insight: Bohr's Atomic Model and Energy Levels

Analyzing the Setup

The Master Equation

Final Calculation

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Bohr’s Quantization for Rotation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Energy of Transition

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Isolating Moment of Inertia

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Substituting Values

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Calculation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Way Forward