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Chapter 6: Problem 98

Find the ionization energy of a hydrogen atom in the fourth energy state.

Short Answer

Expert verified
The ionization energy of a hydrogen atom in the fourth energy state is 0.85 eV.

Step by step solution

01

Find the energy in the fourth energy state

First, we need to find the energy of the hydrogen atom in the fourth energy state (n=4). We'll use the energy level formula for the hydrogen atom: \[E_n = -\dfrac{13.6 \ \text{eV}}{n^2}\] where \(E_n\) is the energy in the n-th energy state, and \(n\) is the principal quantum number. In this case, \(n = 4\). \[E_4 = -\dfrac{13.6 \ \text{eV}}{4^2}\]
02

Calculate the energy in the fourth energy state

Now, we will calculate the energy value: \[E_4 = -\dfrac{13.6 \ \text{eV}}{16} = -0.85 \ \text{eV}\] The energy of the hydrogen atom in the fourth energy state is -0.85 eV.
03

Find ionization energy

The ionization energy is the difference between the energy when the electron is removed from the hydrogen atom (ionized state, \(E_\infty = 0\ \text{eV}\)) and the energy in the fourth energy state, which we have calculated in the previous step as \(E_4\). Ionization energy (IE) = Energy in the ionized state (0 eV) - Energy in the fourth energy state (-0.85 eV) IE = 0 eV - (-0.85 eV)
04

Calculate the ionization energy

Now, we will calculate the ionization energy: IE = 0 eV + 0.85 eV IE = 0.85 eV The ionization energy of a hydrogen atom in the fourth energy state is 0.85 eV.

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