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

Short Answer

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

Step by step solution

01

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).
02

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: \(E_2 = -\frac{13.6\,\text{eV}}{2^2} = -\frac{13.6\,\text{eV}}{4} = -3.4\,\text{eV}\)
03

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 \(0\,\text{eV}\).
04

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 \(3.4\,\text{eV}\).

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Most popular questions from this chapter

An electron in a hydrogen atom is in the ground state (1s). Calculate the probability of finding the electron within a Bohr radius \(\left(a_{0}=0.05295 \mathrm{nm}\right)\) of the proton. The ground state wave function for hydrogen is: \(\psi_{1 s}(r)=A_{1 s} e^{-r / a_{0}}=e^{-r / a_{0}} / \sqrt{\pi a_{0}^{3}}\).

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