The electrons in a `H`- atom kept at rest , jumps from the mth shell to the nth shell `(m gt n)`. Suppose instead of emitting electromagnetic wave, the energy released is converted into kinetic energy of the atom. Assuming Bohr's model and conservation of angular momentum are valid. Now , answer the following questions:
Calculate the angular velocity of the atom about the nucleus if `l` is the momentum of inertia

Options

A. `((m + n))/(6.28) (h)/(l)`
B. `((m + n))/(1.57) (h)/(l)`
C. `((m - n))/(6.28) (h)/(l)`
D. `((m - n))/(1.57) (h)/(l)`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the Concept of Angular Momentum
In a hydrogen atom, the angular momentum \( L \) of an electron in a shell is given by the formula:
\[ L = n \frac{h}{2\pi} \]
where \( n \) is the principal quantum number of the shell, and \( h \) is Planck's constant.

### Step 2: Calculate the Change in Angular Momentum
When the electron jumps from the \( m \)-th shell to the \( n \)-th shell, the change in angular momentum \( \Delta L \) is given by:
\[ \Delta L = L_n - L_m = n \frac{h}{2\pi} - m \frac{h}{2\pi} = (n - m) \frac{h}{2\pi} \]
Since \( m > n \), we can express this as:
\[ \Delta L = (m - n) \frac{h}{2\pi} \]

### Step 3: Relate Angular Momentum to Moment of Inertia and Angular Velocity
According to the conservation of angular momentum, the angular momentum of the atom can also be expressed in terms of its moment of inertia \( I \) and angular velocity \( \omega \):
\[ L = I \omega \]
Given that the moment of inertia is represented as \( l \), we have:
\[ L = l \omega \]

### Step 4: Set the Two Expressions for Angular Momentum Equal
Setting the two expressions for angular momentum equal gives us:
\[ (m - n) \frac{h}{2\pi} = l \omega \]

### Step 5: Solve for Angular Velocity \( \omega \)
Rearranging the equation to solve for \( \omega \):
\[ \omega = \frac{(m - n) \frac{h}{2\pi}}{l} \]
This simplifies to:
\[ \omega = \frac{(m - n) h}{2\pi l} \]

### Final Answer
Thus, the angular velocity \( \omega \) of the atom about the nucleus is:
\[ \omega = \frac{(m - n) h}{2\pi l} \]

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