Question

Ionisation energy of Hydrogen atom is 13.6 eV. In a state where \( n = 2 \), what will be ionisation energy of its electron? 
 

Hint:

The ionization energy decreases with increasing principal quantum number \( n \).

Answer 1

The ionization energy for the \( n \)-th orbit of the hydrogen atom is given by: \[ E_n = \frac{13.6 \, \text{eV}}{n^2}. \] For \( n = 2 \): \[ E_2 = \frac{13.6}{2^2} = \frac{13.6}{4} = 3.4 \, \text{eV}. \] The ionization energy required for ionizing the electron from \( n = 2 \) to \( n = \infty \) is the energy needed to remove it from the \( n = 2 \) level, which is 3.4 eV.