An electron in an 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 reased is converted into the kinetic energy of the atom. Assume the Bohr model and conservation of angular momentum are valid. If `I` is the moment of inertia the angular velocity of the atom about the nucleus is `4 (m - n) h // kI`. Calculate `k`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the Energy Transition
When an electron jumps from the m-th shell to the n-th shell in a hydrogen atom, it releases energy. According to the Bohr model, the energy of an electron in the nth shell is given by:
\[ E_n = -\frac{13.6 \, \text{eV}}{n^2} \]
The energy difference when the electron transitions from the m-th shell to the n-th shell is:
\[ \Delta E = E_n - E_m = -\frac{13.6 \, \text{eV}}{n^2} + \frac{13.6 \, \text{eV}}{m^2} = 13.6 \, \text{eV} \left( \frac{1}{n^2} - \frac{1}{m^2} \right) \]

### Step 2: Convert Energy to Kinetic Energy
The energy released during the transition is converted into kinetic energy of the atom. Thus, we have:
\[ \Delta E = KE = \frac{1}{2} I \omega^2 \]
where \(I\) is the moment of inertia and \(\omega\) is the angular velocity of the atom about the nucleus.

### Step 3: Angular Momentum Conservation
According to the conservation of angular momentum, the change in angular momentum due to the transition can be expressed as:
\[ L_m - L_n = I \omega \]
where \(L_m = \frac{m h}{2 \pi}\) and \(L_n = \frac{n h}{2 \pi}\). Therefore, the change in angular momentum is:
\[ \Delta L = L_m - L_n = \frac{m h}{2 \pi} - \frac{n h}{2 \pi} = \frac{(m - n) h}{2 \pi} \]

### Step 4: Relate Angular Momentum to Angular Velocity
From the conservation of angular momentum, we have:
\[ \Delta L = I \omega \]
Substituting the expression for \(\Delta L\):
\[ \frac{(m - n) h}{2 \pi} = I \omega \]
From this, we can express \(\omega\):
\[ \omega = \frac{(m - n) h}{2 \pi I} \]

### Step 5: Compare with Given Angular Velocity
The problem states that the angular velocity is given by:
\[ \omega = \frac{4(m - n) h}{k I} \]
Setting the two expressions for \(\omega\) equal to each other gives:
\[ \frac{(m - n) h}{2 \pi I} = \frac{4(m - n) h}{k I} \]

### Step 6: Solve for k
Assuming \(m \neq n\) and \(h \neq 0\), we can cancel \((m - n) h\) and \(I\) from both sides:
\[ \frac{1}{2 \pi} = \frac{4}{k} \]
Cross-multiplying gives:
\[ k = 8 \pi \]

### Final Answer
Thus, the value of \(k\) is:
\[ \boxed{8\pi} \]