In a sample of H-atoms , electrons de-excite from a level 'n' to 1 . The total number of lines belonging to Balmer series are two . If the electrons are ionised from level 'n' by photons of energy 13 eV . Then the kinetic energy of the ejected photoelectrons will be :

Options

A. `12.15 eV`
B. `11.49 eV`
C. `12.46 eV`
D. `12.63 eV`

Explanation

To solve the problem, we need to determine the kinetic energy of the ejected photoelectrons when electrons are ionized from a specific energy level in a hydrogen atom. Here’s how we can approach the solution step by step:

### Step 1: Identify the energy level (n)
The problem states that there are two lines in the Balmer series. The Balmer series corresponds to transitions where the final energy level is n=2. The number of lines in the Balmer series can be calculated using the formula for the number of transitions from level n to level 2, which is given by \( \frac{(n-1)(n-2)}{2} \).

Given that there are 2 lines, we can set up the equation:
\[ \frac{(n-1)(n-2)}{2} = 2 \]
Solving this gives:
\[ (n-1)(n-2) = 4 \]
This implies \( n^2 - 3n + 2 = 4 \) or \( n^2 - 3n - 2 = 0 \). Solving this quadratic equation, we find that \( n = 4 \) is the only positive integer solution.

### Step 2: Calculate the energy of the electron at level n=4
The energy of an electron in a hydrogen atom at a given level n is given by the formula:
\[ E_n = -\frac{13.6 \, \text{eV}}{n^2} \]
Substituting \( n = 4 \):
\[ E_4 = -\frac{13.6 \, \text{eV}}{4^2} = -\frac{13.6 \, \text{eV}}{16} = -0.85 \, \text{eV} \]

### Step 3: Determine the energy of the incoming photon
The problem states that the energy of the incoming photon is 13 eV.

### Step 4: Calculate the kinetic energy of the ejected photoelectrons
The kinetic energy (KE) of the ejected photoelectrons can be calculated using the equation:
\[ KE = E_{\text{photon}} - |E_n| \]
Substituting the values we have:
\[ KE = 13 \, \text{eV} - 0.85 \, \text{eV} = 12.15 \, \text{eV} \]

### Conclusion
The kinetic energy of the ejected photoelectrons is \( 12.15 \, \text{eV} \).