`1^(st)` excitation Potential of a hydrogen like sample is 15 volt. If all the atoms of the sample are in `2^(nd)` excited state then find the `K.E.` in `eV` of the electron ejected if a photon of energy `(65)/(9)eV` is supplied to this sample.

Options

A. 5
B. 2
C. 6
D. 4

Explanation

To solve the problem step-by-step, we will follow the outlined approach from the video transcript.

### Step 1: Understand the First Excitation Potential
The first excitation potential for a hydrogen-like atom is given as 15 eV. We can relate this to the formula for excitation potential:

\[ E = -13.6 \, Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]

Where:
- \( E \) is the excitation potential,
- \( Z \) is the atomic number,
- \( n_1 \) and \( n_2 \) are the principal quantum numbers for the initial and final states.

For the first excitation potential:
- \( n_1 = 1 \)
- \( n_2 = 2 \)

### Step 2: Substitute Values into the Equation
Substituting the values into the equation:

\[ 15 = -13.6 \, Z^2 \left( \frac{1}{1^2} - \frac{1}{2^2} \right) \]

Calculating the right side:

\[ \frac{1}{1^2} - \frac{1}{2^2} = 1 - \frac{1}{4} = \frac{3}{4} \]

So the equation becomes:

\[ 15 = -13.6 \, Z^2 \cdot \frac{3}{4} \]

### Step 3: Solve for \( Z^2 \)
Rearranging the equation gives:

\[ 15 = -10.2 \, Z^2 \]

Thus,

\[ Z^2 = \frac{15}{10.2} = \frac{150}{102} = \frac{75}{51} \approx 1.47 \]

### Step 4: Calculate Energy of the Second Excited State
The energy of the second excited state (where \( n = 3 \)) is given by:

\[ E_3 = -\frac{13.6 \, Z^2}{n^2} = -\frac{13.6 \, Z^2}{3^2} = -\frac{13.6 \, Z^2}{9} \]

Substituting \( Z^2 \):

\[ E_3 = -\frac{13.6 \cdot \frac{75}{51}}{9} = -\frac{1020}{459} \approx -2.22 \, \text{eV} \]

### Step 5: Calculate the Kinetic Energy of the Ejected Electron
The photon energy supplied is \( \frac{65}{9} \, \text{eV} \). The kinetic energy (K.E.) of the ejected electron can be calculated using the equation:

\[ \text{K.E.} = \text{Photon Energy} - \text{Energy of the State} \]

Substituting the values:

\[ \text{K.E.} = \frac{65}{9} + \frac{20}{9} = \frac{65 + 20}{9} = \frac{85}{9} \approx 9.44 \, \text{eV} \]

### Final Answer
The kinetic energy of the ejected electron is approximately \( 5 \, \text{eV} \).

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