Question

The ground state energy of the Hydrogen atom is approximately:

• A. \(-1.5\text{ eV}\)
• B. \(-3.4\text{ eV}\)
• C. \(-13.6\text{ eV}\)
• D. \(-54.4\text{ eV}\)

Hint:

The energy levels of hydrogen decrease as \( 1/n^2 \): - Ground state (\( n=1 \)): \( -13.6\text{ eV} \) - First excited state (\( n=2 \)): \( \frac{-13.6}{4} = -3.4\text{ eV} \) - Second excited state (\( n=3 \)): \( \frac{-13.6}{9} \approx -1.51\text{ eV} \)

Answer 1

The Correct Option is C

Concept: The energy levels of an electron bound inside a hydrogen-like atom can be derived using the Bohr model or by solving the Schrödinger equation with a central Coulomb potential. The energy associated with an electron at a principal quantum number level \( n \) is given by: \[ E_n = -\frac{13.6 \cdot Z^2}{n^2}\text{ eV} \] Where:
• \( Z \) is the atomic number of the atom.
• \( n \) is the principal quantum number (\( n = 1, 2, 3, \ldots \)).

Step 1: Identify the specified parameters for the ground state of Hydrogen.

• For a standard Hydrogen atom, the atomic number is \( Z = 1 \).
• The term ground state refers to the lowest possible energy configuration, corresponding to the first orbit level, \( n = 1 \).

Step 2: Calculate the energy value.
Substitute \( Z = 1 \) and \( n = 1 \) into the energy formula: \[ E_1 = -\frac{13.6 \times (1)^2}{(1)^2}\text{ eV} = -13.6\text{ eV} \] The negative sign indicates that the electron is bound within the potential well of the nucleus. To completely remove the electron from its ground state to infinity (where \( E = 0 \)), an ionization energy of \( +13.6\text{ eV} \) must be supplied. This matches Option (C).