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Difficulty: MediumEnergy Levels and Atomic Spectra

The energy of an electron in the first excited state of a hydrogen atom is −3.4 eV-3.4\text{ eV}. What is the minimum energy, in joules (J\text{J}), required to completely remove the electron from this state to ionize the atom? (1 eV=1.6×10−19 J1\text{ eV} = 1.6 \times 10^{-19}\text{ J})

Answer: 5.44e-19 J

Answer

The minimum energy required to ionize the atom from its first excited state is 5.44×10−19 J5.44 \times 10^{-19}\text{ J}.
Ionization energy is defined as the minimum energy necessary to completely remove an electron from its bound atomic energy level to infinity (E∞=0 eVE_{\infty} = 0\text{ eV}). For an electron at E=−3.4 eVE = -3.4\text{ eV}, the required energy change is ΔE=0−(−3.4 eV)=3.4 eV\Delta E = 0 - (-3.4\text{ eV}) = 3.4\text{ eV}. Converting this to joules using 1 eV=1.6×10−19 J1\text{ eV} = 1.6 \times 10^{-19}\text{ J} gives 3.4×1.6×10−19=5.44×10−19 J3.4 \times 1.6 \times 10^{-19} = 5.44 \times 10^{-19}\text{ J}.

Step-by-Step Solution

1
Determine the energy required for ionization in electron-volts (eV)
ΔE=E∞−E2=0 eV−(−3.4 eV)=3.4 eV\Delta E = E_{\infty} - E_2 = 0\text{ eV} - (-3.4\text{ eV}) = 3.4\text{ eV}
Ionization requires supplying sufficient energy to raise the electron from its bound energy state (E2=−3.4 eVE_2 = -3.4\text{ eV}) to the ionization limit where it is free (E∞=0 eVE_{\infty} = 0\text{ eV}).
2
Convert the ionization energy from electron-volts to joules
E=3.4 eV×1.6×10−19 J/eV=5.44×10−19 JE = 3.4\text{ eV} \times 1.6 \times 10^{-19}\text{ J/eV} = 5.44 \times 10^{-19}\text{ J}
To convert energy from electron-volts (eV) to joules (J), multiply the value in eV by the elementary charge conversion factor 1.6×10−19 J/eV1.6 \times 10^{-19}\text{ J/eV}.

Key Concept

Ionization Energy and Energy Level Transitions
Estimated Time:1m 30s
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Topics

Energy Levels and Atomic Spectra

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