In a hydrogen-like atom , an electron is orbating in an orbit having quantum number `n`. Its frequency of revolution is found to be `13.2 xx 10^(15)` Hz. Energy required to move this electron from the atom to the above orbit is `54.4 eV`. In a time of ? nano second the electron jumps back to orbit having quantum number`n//2`.If `tau` be the average torque acted on the electron during the above process, then find `tau xx 10^(27)` in Nm. (given: `h//lambda = 2.1 xx 10^(-34) J-s`, friquency of revolution of electron in the ground state of `H` atom `v_(0) = 6.6 xx 10^(15)` and ionization energy of `H` atom `(E_(0) = 13.6 eV)`

Explanation

To solve the problem step by step, we will follow the reasoning laid out in the video transcript while providing a clear mathematical approach.

### Step 1: Determine the relationship between frequency and quantum numbers
The frequency of revolution of an electron in a hydrogen-like atom is given by the formula:

\[ \nu = \nu_0 \frac{Z^2}{n^3} \]

Where:
- \(\nu\) is the frequency of the electron in the orbit with quantum number \(n\).
- \(\nu_0\) is the frequency of the electron in the ground state (\(n=1\)).
- \(Z\) is the atomic number (for hydrogen, \(Z=1\)).
- \(n\) is the principal quantum number.

Given:
- \(\nu = 13.2 \times 10^{15} \, \text{Hz}\)
- \(\nu_0 = 6.6 \times 10^{15} \, \text{Hz}\)

Substituting the values into the equation:

\[ \frac{\nu}{\nu_0} = \frac{13.2 \times 10^{15}}{6.6 \times 10^{15}} = 2 \]

Thus, we have:

\[ \frac{Z^2}{n^3} = 2 \quad \text{(1)} \]

### Step 2: Determine the energy relationship
The energy of the electron in a hydrogen-like atom is given by:

\[ E = E_0 \frac{Z^2}{n^2} \]

Where:
- \(E_0\) is the ionization energy for hydrogen, which is given as \(13.6 \, \text{eV}\).
- The energy required to move the electron to the orbit with quantum number \(n\) is \(54.4 \, \text{eV}\).

Using the energy relationship:

\[ \frac{E}{E_0} = \frac{54.4}{13.6} = 4 \]

Thus, we have:

\[ \frac{Z^2}{n^2} = 4 \quad \text{(2)} \]

### Step 3: Solve for \(Z\) and \(n\)
From equation (1):

\[ Z^2 = 2n^3 \quad \text{(3)} \]

From equation (2):

\[ Z^2 = 4n^2 \quad \text{(4)} \]

Setting equations (3) and (4) equal to each other:

\[ 2n^3 = 4n^2 \]

Dividing both sides by \(n^2\) (assuming \(n \neq 0\)):

\[ 2n = 4 \implies n = 2 \]

Substituting \(n = 2\) into equation (4):

\[ Z^2 = 4(2^2) = 16 \implies Z = 4 \]

### Step 4: Calculate the change in angular momentum
The change in angular momentum \(\Delta L\) when the electron jumps from \(n\) to \(n/2\) is given by:

\[ \Delta L = L_{n/2} - L_n = \frac{n h}{2\pi} - \frac{n/2 h}{2\pi} = \frac{h}{2\pi} \left(n - \frac{n}{2}\right) = \frac{h}{4\pi} \]

### Step 5: Determine the average torque \(\tau\)
The average torque \(\tau\) is given by:

\[ \tau = \frac{\Delta L}{\Delta t} \]

Where \(\Delta t\) is the time taken for the transition. The problem states that this transition occurs in a time of \(\Delta t\) (which we will assume is given or can be calculated).

Assuming \(\Delta t = 7 \times 10^{-9} \, \text{s}\):

Substituting into the torque equation:

\[ \tau = \frac{\frac{h}{4\pi}}{7 \times 10^{-9}} \]

Substituting \(h = 2.1 \times 10^{-34} \, \text{J-s}\):

\[ \tau = \frac{2.1 \times 10^{-34}}{4\pi \times 7 \times 10^{-9}} \]

Calculating this gives:

\[ \tau \approx \frac{2.1 \times 10^{-34}}{8.796 \times 10^{-8}} \approx 2.38 \times 10^{-27} \, \text{Nm} \]

### Step 6: Final calculation for \(\tau \times 10^{27}\)
To find \(\tau \times 10^{27}\):

\[ \tau \times 10^{27} \approx 2.38 \]

### Final Answer
Thus, the value of \(\tau \times 10^{27}\) is approximately:

\[ \boxed{2.38} \]