The ionization potential of the hydrogen atom is 13.6 V. The energy needed to ionize a hydrogen atom which is in its first excited state is about

Options

A. 13.6 eV
B. 10.2 eV
C. 3.4 eV
D. 1.5 eV

Answer

To find the energy needed to ionize a hydrogen atom that is in its first excited state, we can follow these steps:

### Step 1: Understand the Ionization Potential
The ionization potential of a hydrogen atom in its ground state (n=1) is given as 13.6 eV. This means that to remove an electron from the ground state, we need to supply 13.6 eV of energy.

### Step 2: Determine the Energy of the First Excited State
The energy of an electron in a hydrogen atom at a given principal quantum number \( n \) is given by the formula:
\[ E_n = -\frac{13.6 \, \text{eV}}{n^2} \]
For the first excited state, \( n = 2 \):
\[ E_2 = -\frac{13.6 \, \text{eV}}{2^2} = -\frac{13.6 \, \text{eV}}{4} = -3.4 \, \text{eV} \]

### Step 3: Calculate the Energy Needed for Ionization
To ionize the hydrogen atom from the first excited state, we need to supply enough energy to overcome the negative energy of the state and bring the electron to zero energy (free state). The energy needed to ionize the atom is given by:
\[ \text{Energy needed} = 0 \, \text{eV} - E_2 \]
Substituting the value of \( E_2 \):
\[ \text{Energy needed} = 0 \, \text{eV} - (-3.4 \, \text{eV}) = 3.4 \, \text{eV} \]

### Final Answer
The energy needed to ionize a hydrogen atom which is in its first excited state is **3.4 eV**.

---