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 start by understanding the relationship between the ionization potential of a hydrogen atom and the energy required to remove an electron from a specific orbit.

### Step 1: Understand the Ionization Potential
The ionization potential of hydrogen is given as \( 13.6 \, \text{eV} \). This is the energy required to remove an electron from the ground state (first orbit, \( n=1 \)) of the hydrogen atom.

### Step 2: Use the Formula for Energy Levels
The energy of an electron in the \( n \)-th orbit of a hydrogen atom is given by the formula:
\[ E_n = -\frac{13.6 \, \text{eV}}{n^2} \]
where \( n \) is the principal quantum number.

### Step 3: Calculate the Energy for the Third Orbit
For the third orbit (\( n = 3 \)):
\[ E_3 = -\frac{13.6 \, \text{eV}}{3^2} = -\frac{13.6 \, \text{eV}}{9} = -1.511 \, \text{eV} \]
This value is negative because it indicates that the electron is bound to the nucleus.

### Step 4: Calculate the Energy Required to Remove the Electron
The energy required to remove the electron from the third orbit (ionization energy) is the positive value of \( E_3 \):
\[ \text{Energy required} = -E_3 = 1.511 \, \text{eV} \]

### Step 5: Relate to \( k/2 \)
According to the problem, the energy required to remove the electron from the third orbit is given as \( \frac{k}{2} \, \text{eV} \). Therefore, we can set up the equation:
\[ \frac{k}{2} = 1.511 \, \text{eV} \]

### Step 6: Solve for \( k \)
To find \( k \), we multiply both sides of the equation by 2:
\[ k = 2 \times 1.511 = 3.022 \]

Thus, the value of \( k \) is:
\[ \boxed{3.022} \]