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

Answer

To solve the problem step by step, we need to follow these calculations:

### Step 1: Determine the Energy Level (n)
The problem states that the total number of lines in the Balmer series is two. The Balmer series corresponds to transitions from higher energy levels (n > 2) to n = 2. The two lines indicate that the electrons are transitioning from n = 3 and n = 4 to n = 2. Therefore, the highest energy level from which the electrons are de-exciting is n = 4.

### Step 2: Calculate the Energy of the Electron in the n = 4 Level
The energy of an electron in the nth level of a hydrogen atom is given by the formula:
\[ E_n = -\frac{13.6 \, \text{eV}}{n^2} \]
For 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 Required for Ionization
To ionize the electron from the n = 4 level, we need to provide energy equal to the absolute value of the energy at that level:
\[ \text{Ionization Energy} = 0.85 \, \text{eV} \]

### Step 4: Calculate the Kinetic Energy of the Ejected Photoelectrons
The energy of the photon used for ionization is given as 13 eV. The kinetic energy (KE) of the ejected photoelectrons can be calculated using the formula:
\[ \text{Photon Energy} = \text{Ionization Energy} + \text{Kinetic Energy} \]
Rearranging gives:
\[ \text{Kinetic Energy} = \text{Photon Energy} - \text{Ionization Energy} \]
Substituting the known values:
\[ \text{Kinetic Energy} = 13 \, \text{eV} - 0.85 \, \text{eV} = 12.15 \, \text{eV} \]

### Final Answer
The kinetic energy of the ejected photoelectrons will be **12.15 eV**.

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