Photons of equal energy were incident on two different gas samples. One sample containing H-atoms in the ground state and the other sample containing H-atoms in some excited state with a principle quantum number 'n'. The photonic beams totally ionise the H-atoms. If the difference in the kinetic energy of the ejected electrons in the two different cases is `12.75 eV`. Then find the principal quantum number 'n' of the excited state.

Explanation

To solve the problem, we need to analyze the kinetic energy of the electrons ejected from hydrogen atoms in two different states: the ground state (n=1) and an excited state (n). We are given that the difference in kinetic energy of the ejected electrons is 12.75 eV. Let's break down the solution step by step.

### Step 1: Understanding Kinetic Energy in Both Cases
For the hydrogen atom in the ground state (n=1), the kinetic energy (KE1) of the ejected electron can be expressed as:
\[ KE_1 = E_{\text{photon}} - BE_1 \]
where \( BE_1 \) is the binding energy of the electron in the ground state.

For the hydrogen atom in the excited state (n), the kinetic energy (KE2) can be expressed as:
\[ KE_2 = E_{\text{photon}} - BE_n \]
where \( BE_n \) is the binding energy of the electron in the excited state.

### Step 2: Binding Energy Formula
The binding energy for a hydrogen atom in a state with principal quantum number n is given by:
\[ BE_n = -\frac{13.6 \, \text{eV}}{n^2} \]
Thus, for n=1:
\[ BE_1 = -\frac{13.6 \, \text{eV}}{1^2} = -13.6 \, \text{eV} \]
And for the excited state n:
\[ BE_n = -\frac{13.6 \, \text{eV}}{n^2} \]

### Step 3: Setting Up the Equation for Kinetic Energy Difference
The difference in kinetic energy is given as:
\[ KE_2 - KE_1 = 12.75 \, \text{eV} \]
Substituting the expressions for KE1 and KE2:
\[ (E_{\text{photon}} - BE_n) - (E_{\text{photon}} - BE_1) = 12.75 \]
This simplifies to:
\[ BE_1 - BE_n = 12.75 \]

### Step 4: Plugging in the Binding Energy Values
Substituting the binding energy values:
\[ -(-13.6) - \left(-\frac{13.6}{n^2}\right) = 12.75 \]
This simplifies to:
\[ 13.6 + \frac{13.6}{n^2} = 12.75 \]

### Step 5: Rearranging the Equation
Rearranging gives:
\[ \frac{13.6}{n^2} = 12.75 - 13.6 \]
\[ \frac{13.6}{n^2} = -0.85 \]
This leads to:
\[ 13.6 = -0.85n^2 \]
However, since binding energy cannot be negative, we need to correct the equation:
\[ 13.6 - 12.75 = \frac{13.6}{n^2} \]
\[ 0.85 = \frac{13.6}{n^2} \]

### Step 6: Solving for n^2
Now, we can solve for \( n^2 \):
\[ n^2 = \frac{13.6}{0.85} \]
Calculating this gives:
\[ n^2 \approx 16 \]

### Step 7: Finding n
Taking the square root:
\[ n = 4 \]

### Final Answer
The principal quantum number \( n \) of the excited state is:
\[ \boxed{4} \]