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 7 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 information provided in the question and the video transcript.

### 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:

\[ v = v_0 \cdot \frac{Z^2}{n^3} \]

Where:
- \( v \) is the frequency of the electron in the orbit with quantum number \( n \).
- \( v_0 \) is the frequency of the electron in the ground state (for hydrogen, \( v_0 = 6.6 \times 10^{15} \) Hz).
- \( Z \) is the atomic number.
- \( n \) is the principal quantum number.

Given:
- \( v = 13.2 \times 10^{15} \) Hz

Substituting the known values:

\[ 13.2 \times 10^{15} = 6.6 \times 10^{15} \cdot \frac{Z^2}{n^3} \]

### Step 2: Simplify the equation
Dividing both sides by \( 6.6 \times 10^{15} \):

\[ \frac{13.2}{6.6} = \frac{Z^2}{n^3} \]

Calculating \( \frac{13.2}{6.6} = 2 \):

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

### Step 3: Relate energy and quantum numbers
The energy of an electron in a hydrogen-like atom is given by:

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

Where:
- \( E_0 = 13.6 \) eV (ionization energy of hydrogen).

Given that the energy required to move the electron to the orbit is \( 54.4 \) eV:

\[ 54.4 = 13.6 \cdot \frac{Z^2}{n^2} \]

### Step 4: Simplify the energy equation
Dividing both sides by \( 13.6 \):

\[ \frac{54.4}{13.6} = \frac{Z^2}{n^2} \]

Calculating \( \frac{54.4}{13.6} = 4 \):

\[ 4 = \frac{Z^2}{n^2} \]

### Step 5: Solve for \( Z^2 \) and \( n \)
From the two equations we have:

1. \( Z^2 = 2n^3 \)
2. \( Z^2 = 4n^2 \)

Setting them 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 \( Z^2 = 4n^2 \):

\[ Z^2 = 4 \cdot 2^2 = 16 \implies Z = 4 \]

### Step 6: Calculate the torque
The angular momentum \( L \) of the electron is given by:

\[ L = \frac{nh}{2\pi} \]

Differentiating with respect to time gives us the torque \( \tau \):

\[ \tau = \frac{dL}{dt} = \frac{h}{2\pi} \cdot \frac{dn}{dt} \]

The change in quantum number \( dn = n - \frac{n}{2} = 2 - 1 = 1 \).

Given the time \( dt = 7 \) ns = \( 7 \times 10^{-9} \) s:

\[ \frac{dn}{dt} = \frac{1}{7 \times 10^{-9}} \]

Substituting into the torque equation:

\[ \tau = \frac{h}{2\pi} \cdot \frac{1}{7 \times 10^{-9}} \]

Substituting \( h = 6.63 \times 10^{-34} \) J·s:

\[ \tau = \frac{6.63 \times 10^{-34}}{2\pi} \cdot \frac{1}{7 \times 10^{-9}} \]

Calculating \( \tau \):

\[ \tau \approx \frac{6.63 \times 10^{-34}}{6.2832 \times 7 \times 10^{-9}} \approx \frac{6.63 \times 10^{-34}}{4.398 \times 10^{-8}} \approx 1.5 \times 10^{-26} \text{ Nm} \]

### Step 7: Final calculation for \( \tau \times 10^{27} \)
\[ \tau \times 10^{27} \approx 1.5 \times 10^{-26} \times 10^{27} = 15 \]

### Final Answer:
\[ \tau \times 10^{27} \approx 15 \text{ Nm} \]