How to Calculate the Ionization Energy of Hydrogen

The ionization energy of hydrogen is the energy needed to remove the electron from an isolated hydrogen atom in its ground state. You can calculate it as the energy gap between the lowest bound level and the ionization limit. The standard result is about 13.6 eV per atom, or 1312 kJ mol⁻¹.

Use the hydrogen energy levels

In the Bohr model, the energy of a hydrogen atom in level $n$ is

$$E_n=-\frac{hcR_H}{n^2},$$

where $h$ is Planck’s constant, $c$ is the speed of light, and $R_H$ is the Rydberg constant for hydrogen. The negative sign indicates that the electron is bound to the proton. The ionization limit is $n=\infty$, for which the energy is zero: $E_\infty=0$.

For a ground-state atom, $n=1$. The energy that must be supplied is therefore

$$\begin{aligned} \Delta E &= E_\infty-E_1 \\ &=0-(-hcR_H) \\ &=hcR_H. \end{aligned}$$

The energy is positive because the atom must absorb energy to reach the unbound state. The corresponding spectral wavenumber is $R_H\approx1.0967758\times10^7\ \text{m}^{-1}$.

Worked calculation

Substitute the constants into $\Delta E=hcR_H$:

$$\begin{aligned} \Delta E &\approx (6.62607015\times10^{-34}\ \text{J s}) (2.99792458\times10^8\ \text{m s}^{-1}) (1.0967758\times10^7\ \text{m}^{-1}) \\ &\approx 2.1787\times10^{-18}\ \text{J per atom}. \end{aligned}$$

To express this in electronvolts, divide by the elementary charge, $1.602176634\times10^{-19}$ joules per electronvolt:

$$\Delta E\approx\frac{2.1787\times10^{-18}}{1.602176634\times10^{-19}} \approx13.60\ \text{eV per atom}.$$

To convert to energy per mole, multiply by Avogadro’s constant, $N_A=6.02214076\times10^{23}\ \text{mol}^{-1}$:

$$\begin{aligned} \Delta E_{\text{mol}}&\approx(2.1787\times10^{-18})(6.02214076\times10^{23})\ \text{J mol}^{-1}\\ &\approx1.312\times10^6\ \text{J mol}^{-1}\\ &\approx1312\ \text{kJ mol}^{-1}. \end{aligned}$$

Thus the same ionization energy can be stated as approximately 13.6 eV per atom, 2.18 × 10⁻¹⁸ J per atom, or 1312 kJ mol⁻¹. A high-precision spectroscopic value is about 13.5984 eV per atom; the extra digits are not implied by the simplified calculation above.

What the number means

The ionization energy is a threshold: an electron must receive at least this much energy to escape from a ground-state hydrogen atom. A photon at the threshold has wavelength about $1/R_H\approx91.2\ \text{nm}$, in the ultraviolet. A photon with a shorter wavelength has more energy and can ionize the atom; a longer-wavelength photon does not have enough energy in a single-photon process.

For an atom already in an excited level, the required energy is lower. From level $n$, the idealized hydrogen-model result is $hcR_H/n^2$. For example, from $n=2$ it is one quarter of the ground-state value, about 3.40 eV. In chemistry tables, “the ionization energy of hydrogen” ordinarily means removal of an electron from ground-state gaseous H atoms, not from an excited atom or a hydrogen molecule.

Common mistakes

• Keeping the negative sign as the answer: $E_1$ is negative because it is a bound-state energy. The energy required to ionize is the positive difference $E_\infty-E_1$.
• Stopping at joules per atom: If a question asks for kJ mol⁻¹, multiply by Avogadro’s constant and divide by 1000 to convert joules to kilojoules.
• Mixing up eV per atom and kJ mol⁻¹: These are different units for the same physical energy, but the conversion includes both the electronvolt-to-joule factor and Avogadro’s constant. One eV per particle corresponds to about 96.485 kJ mol⁻¹.
• Using the wrong starting state: The 13.6 eV value applies to ground-state atomic hydrogen. Excited hydrogen needs less energy to ionize, while molecular hydrogen has a different ionization energy.
• Expecting every model to give identical final digits: The elementary Bohr calculation is an approximation. More precise values account for effects beyond that simple model; typical chemistry problems round the result to 13.6 eV or 1312 kJ mol⁻¹.