Question

If the binding energy of ground state electron in a hydrogen atom is $136 eV$, then, the energy required to remove the electron from the second excited state of $Li ^{2+}$ will be : $x \times 10^{-1} eV$. The value of $x$ is

Hint:

The energy required to ionize an electron in a hydrogen-like atom depends on the atomic number \( Z \) and the principal quantum number \( n \).

Answer 1

Correct Answer: 136

The energy levels in a hydrogen-like atom are given by the formula: \[ E_n = -13.6 \times \frac{Z^2}{n^2} \, \text{eV} \] where \( Z \) is the atomic number and \( n \) is the principal quantum number. For \( \text{Li}^{2+} \), \( Z = 3 \), and the second excited state corresponds to \( n = 3 \). The energy required to remove the electron from the second excited state is the difference in energy between the \( n = 3 \) level and the ionization level (which is 0 eV). Therefore: \[ E_3 = -13.6 \times \frac{3^2}{3^2} = -13.6 \, \text{eV} \] The energy required to remove the electron from the second excited state is: \[ |E_3| = 13.6 \, \text{eV} \] Thus, \( x = 136 \).