Question

What is the ionization energy of a hydrogen atom excited to the \(n=2\) state?

Short answer

Answer: The ionization energy of a hydrogen atom excited to the \(n=2\) state is \(3.4\,\text{eV}\).

Step-by-step solution

Identify the hydrogen atomic energy level equation

The energy levels of a hydrogen atom can be calculated using the following equation: \(E_n = -\frac{13.6\,\text{eV}}{n^2}\) where \(n\) is the principal quantum number, and the energy \(E_n\) is in electron volts (eV).

Calculate the energy of the hydrogen atom at the \(n=2\) state

Using the energy level equation, we can calculate the energy of the hydrogen atom at the \(n=2\) state: \(E_2 = -\frac{13.6\,\text{eV}}{2^2} = -\frac{13.6\,\text{eV}}{4} = -3.4\,\text{eV}\)

Calculate the energy of the ionized state

When the hydrogen atom is ionized, the electron is completely removed and the energy of the atom is \(0\,\text{eV}\).

Calculate the ionization energy

The ionization energy is the difference in energy between the \(n=2\) state and the ionized state: Ionization Energy = \(E_\text{ionized} - E_2 = 0\,\text{eV} - (-3.4\,\text{eV}) = 3.4\,\text{eV}\) So, the ionization energy of a hydrogen atom excited to the \(n=2\) state is \(3.4\,\text{eV}\).