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

To determine the energy required to remove an electron from the second excited state of Li²⁺, we use the expression for the energy levels of hydrogen-like atoms: \( E_n = -\frac{Z^2 \cdot 13.6\; \text{eV}}{n^2} \), where \( Z \) is the atomic number, and \( n \) is the principal quantum number.

For a hydrogen atom, the ground state energy is given as 136 eV. However, this seems to be a typographical error as the actual value is 13.6 eV. Assuming a typo, let us proceed with 13.6 eV:

• The atomic number \( Z \) for Li²⁺ is 3.
• The second excited state corresponds to \( n = 3 \).

Substituting the values into the energy level equation:

Energy, \( E_3 = -\frac{3^2 \cdot 13.6}{3^2} = -13.6\; \text{eV} \).

The energy required to ionize the atom (i.e., remove the electron) is the absolute value of this energy level:

\( |E_3| = 13.6\; \text{eV} \).

Given the problem states that this energy corresponds to \( x \times 10^{-1}\; \text{eV} \), we equate:

\( x \times 10^{-1} = 13.6 \)

Solving for \( x \), we find:

\( x = 136 \)

Thus, the value of \( x \) is 136, which is within the expected range.