The ionisation potential of hydrogen atom is `13.6 eV` The energy required to remve as electron in the `n = 2` state of the hydrogen atom is

Options

A. `3.4 eV`
B. `6.8 eV`
C. `13.6 eV`
D. `27.7 eV`

Answer

To find the energy required to remove an electron from the n = 2 state of the hydrogen atom, we can follow these steps:

### Step 1: Understand Ionization Energy
The ionization energy (or ionization potential) is the minimum energy required to remove an electron from an atom. For hydrogen, the ionization potential from the ground state (n = 1) is given as 13.6 eV.

### Step 2: Calculate the Energy of the Electron in the n = 2 State
The energy of an electron in a hydrogen atom can be calculated using the formula:
\[ E_n = -\frac{13.6 \, \text{eV} \cdot Z^2}{n^2} \]
For hydrogen, the atomic number \( Z = 1 \) and for the n = 2 state, \( n = 2 \).

Substituting the values:
\[ E_2 = -\frac{13.6 \, \text{eV} \cdot 1^2}{2^2} \]
\[ E_2 = -\frac{13.6 \, \text{eV}}{4} \]
\[ E_2 = -3.4 \, \text{eV} \]

### Step 3: Calculate the Ionization Energy from the n = 2 State
The energy required to remove the electron from the n = 2 state to infinity (ionization) is given by:
\[ \text{Ionization Energy} = 0 - E_2 \]
Substituting the value we found:
\[ \text{Ionization Energy} = 0 - (-3.4 \, \text{eV}) \]
\[ \text{Ionization Energy} = 3.4 \, \text{eV} \]

### Final Answer
The energy required to remove an electron in the n = 2 state of the hydrogen atom is **3.4 eV**.

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