If the ionization potential of hydrogen atom is `13.6eV`, the energy required to remove from the third orbit of hydrogen atom is `k//2eV`. Find the value of `k`…….

Explanation

To find the value of \( k \) in the given problem, we will follow these steps:

### Step 1: Understand the Ionization Energy Formula
The energy required to remove an electron from an orbit in a hydrogen atom is given by the formula:

\[ E = \frac{E_0}{n^2} \]

where:
- \( E \) is the energy required to remove the electron from the nth orbit,
- \( E_0 \) is the ionization potential of hydrogen (given as \( 13.6 \, \text{eV} \)),
- \( n \) is the principal quantum number (for the third orbit, \( n = 3 \)).

### Step 2: Substitute Known Values
Substituting the known values into the formula:

\[ E = \frac{13.6 \, \text{eV}}{3^2} = \frac{13.6 \, \text{eV}}{9} \]

### Step 3: Calculate the Energy
Now, we calculate the energy:

\[ E = \frac{13.6}{9} \, \text{eV} \approx 1.5111 \, \text{eV} \]

### Step 4: Relate to Given Expression
According to the problem, the energy required to remove the electron from the third orbit is also given as:

\[ E = \frac{k}{2} \, \text{eV} \]

### Step 5: Set Up the Equation
Now we equate the two expressions for energy:

\[ \frac{k}{2} = \frac{13.6}{9} \]

### Step 6: Solve for \( k \)
To find \( k \), we multiply both sides by 2:

\[ k = 2 \times \frac{13.6}{9} \]

Calculating this gives:

\[ k = \frac{27.2}{9} \approx 3.0222 \]

### Step 7: Round to Nearest Integer
Since \( k \) is often expressed as an integer in physics problems, we can round this value:

\[ k \approx 3 \]

### Final Answer
Thus, the value of \( k \) is:

\[ \boxed{3} \]

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