The ionisation energy of electron (E) is second orbit of hydrogen atom is

Options

A. -13.6 eV
B. -3.4 eV
C. +3.4 eV
D. zero

Answer

To find the ionization energy of the electron in the second orbit of a hydrogen atom, we can follow these steps:

### Step-by-Step Solution:

1. **Understand Ionization Energy**:
Ionization energy is defined as the amount of energy required to remove an electron from an atom. For hydrogen, this energy can be calculated using the formula derived from Bohr's model.

2. **Bohr's Formula for Energy Levels**:
The energy of an electron in the nth orbit of a hydrogen atom is given by the formula:
\[ E_n = -\frac{13.6 \, \text{eV} \cdot Z^2}{n^2} \]
where \( Z \) is the atomic number and \( n \) is the principal quantum number (the orbit number).

3. **Identify Values for Hydrogen**:
For hydrogen:
- The atomic number \( Z = 1 \).
- We are looking for the energy in the second orbit, so \( n = 2 \).

4. **Substitute Values into the Formula**:
Plugging in the values into the formula:
\[ E_2 = -\frac{13.6 \, \text{eV} \cdot 1^2}{2^2} \]
This simplifies to:
\[ E_2 = -\frac{13.6 \, \text{eV}}{4} \]

5. **Calculate the Energy**:
Performing the division:
\[ E_2 = -3.4 \, \text{eV} \]

6. **Determine Ionization Energy**:
Since ionization energy is the energy required to remove the electron from the atom, we take the absolute value of \( E_2 \):
\[ \text{Ionization Energy} = 3.4 \, \text{eV} \]

### Final Answer:
The ionization energy of the electron in the second orbit of a hydrogen atom is **3.4 eV**.

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