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 an electron in the second orbit of a hydrogen atom, we can follow these steps:

### Step 1: Understand the formula for energy levels in hydrogen
The energy of an electron in the nth orbit of a hydrogen atom is given by the formula:
\[ E_n = -\frac{13.6 \, Z^2}{n^2} \, \text{eV} \]
where:
- \( E_n \) is the energy of the electron in the nth orbit,
- \( Z \) is the atomic number (for hydrogen, \( Z = 1 \)),
- \( n \) is the principal quantum number (the orbit number).

### Step 2: Substitute the values into the formula
For the second orbit, \( n = 2 \) and \( Z = 1 \):
\[ E_2 = -\frac{13.6 \times 1^2}{2^2} = -\frac{13.6}{4} = -3.4 \, \text{eV} \]

### Step 3: Understand ionization energy
Ionization energy is defined as the energy required to remove an electron completely from the atom. Since the energy of the electron in the second orbit is negative, it means that the electron is bound to the nucleus and energy must be supplied to remove it.

### Step 4: Calculate the ionization energy
To ionize the electron from the second orbit, we need to provide energy equal to the absolute value of the energy of the electron in that orbit:
\[ \text{Ionization Energy} = |E_2| = 3.4 \, \text{eV} \]

### Conclusion
The ionization energy of an electron in the second orbit of a hydrogen atom is:
\[ \text{Ionization Energy} = 3.4 \, \text{eV} \]

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