An electron moving along `x` -axis is confined between two walls at `x=0` and at `x=L`. There is no loss of energy during collision between the electron and the wall. Initially, the electron is in quantum state `n=2` and then it jumps to quantumi state `n=1`. The emitted photon is incident on a metal having work function equal to energy of electron in quantum state `n=1`. If maximum possible kinetic energy of emitted photoelectron is `x` times the energy of electron in quantum state `n=1`, then find `x`.

Answer

To solve the problem, we will follow these steps:

### Step 1: Determine the energy of the electron in quantum states
The energy of an electron in a one-dimensional box (particle in a box) is given by the formula:
\[ E_n = \frac{n^2 h^2}{8mL^2} \]
where:
- \(E_n\) is the energy of the electron in the \(n^{th}\) quantum state,
- \(h\) is Planck's constant,
- \(m\) is the mass of the electron,
- \(L\) is the length of the box,
- \(n\) is the quantum number.

For \(n = 1\):
\[ E_1 = \frac{1^2 h^2}{8mL^2} = \frac{h^2}{8mL^2} \]

For \(n = 2\):
\[ E_2 = \frac{2^2 h^2}{8mL^2} = \frac{4h^2}{8mL^2} = \frac{h^2}{2mL^2} \]

### Step 2: Calculate the energy difference when the electron transitions
When the electron transitions from \(n = 2\) to \(n = 1\), the energy difference (which corresponds to the energy of the emitted photon) is:
\[ E_{\text{photon}} = E_2 - E_1 = \frac{h^2}{2mL^2} - \frac{h^2}{8mL^2} \]
To simplify:
\[ E_{\text{photon}} = \frac{4h^2}{8mL^2} - \frac{h^2}{8mL^2} = \frac{3h^2}{8mL^2} \]

### Step 3: Relate the emitted photon energy to the work function
According to the problem, the emitted photon is incident on a metal with a work function equal to the energy of the electron in quantum state \(n = 1\):
\[ \phi = E_1 = \frac{h^2}{8mL^2} \]

### Step 4: Apply the photoelectric effect equation
Using the photoelectric effect equation:
\[ E_{\text{photon}} = K.E + \phi \]
Where \(K.E\) is the kinetic energy of the emitted photoelectron. Substituting the values we have:
\[ \frac{3h^2}{8mL^2} = K.E + \frac{h^2}{8mL^2} \]
Rearranging gives:
\[ K.E = \frac{3h^2}{8mL^2} - \frac{h^2}{8mL^2} = \frac{2h^2}{8mL^2} = \frac{h^2}{4mL^2} \]

### Step 5: Relate the kinetic energy to the energy in state \(n = 1\)
We know:
\[ E_1 = \frac{h^2}{8mL^2} \]
Thus, we can express the kinetic energy as:
\[ K.E = 2E_1 \]

### Step 6: Find the value of \(x\)
From the problem statement, we have:
\[ K.E = x \cdot E_1 \]
Substituting the expression we found:
\[ 2E_1 = x \cdot E_1 \]
Dividing both sides by \(E_1\) (assuming \(E_1 \neq 0\)):
\[ x = 2 \]

### Final Answer
Thus, the value of \(x\) is:
\[ \boxed{2} \]