Question

calculate ionisation energy of hydrogen atom in kJ mol1

Answer 1

Answer on Question #66834 – Chemistry | Inorganic Chemistry

Calculate ionization energy of hydrogen atom in $\mathrm{kJ}^{*}\mathrm{mol}^{-1}$.

Solution:

Use the Rydberg expression:

The wavelength $\lambda$ of the emission line in the hydrogen spectrum is given by:

$$\frac{1}{\lambda} = R \left( \frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}} \right)$$

R is the Rydberg Constant and has the value $1.097 \times 10^{7} \mathrm{~m}^{-1}$

$n_1$ is the principle quantum number of the lower energy level

$n_2$ is the principle quantum number of the higher energy level.

The energy levels in hydrogen converge and coalesce:

Energy-level diagram for hydrogen

© 2009 Encyclopædia Britannica, Inc.

Since the electron is in the $n_1 = 1$ ground state we need to consider series 1. These transitions occur in the UV - part of the spectrum and are known as The Lyman Series.

You can see that as the value of $n_2$ increases then the value of $1 / n_2^2$ decreases. At higher and higher values the expression tends to zero until at $n = \infty$ we can consider the electron to have left the atom resulting in an ion.

The Rydberg expression now becomes:

$$\frac{1}{\lambda} = R \left( \frac{1}{n_{1}^{2}} - 0 \right)$$

Since $n_1 = 1$ this becomes:

$$\frac{1}{\lambda} = R = 1.097 * 10^{7} \Rightarrow \lambda = 9.116 * 10^{-8} \, (m)$$

We can now find the frequency and hence the corresponding energy:

$$c = \nu * \lambda \Rightarrow \nu = c / \lambda$$ $$\nu = \frac {c}{\lambda} = \frac {3 * 10^{8}}{9.116 * 10^{-8}} = 3.291 * 10^{15} \ (s^{-1})$$

Now we can use the Planck expression:

$$E = h * v$$ $$E = 6.626 * 10^{-34} * 3.291 * 10^{15} = 2.18 * 10^{-18} \ (J)$$

This is the energy needed to remove 1 electron from 1 hydrogen atom. To find the energy required to ionize 1 mole of H atoms we multiply by the Avogadro Constant:

$$E = 2.18 * 10^{-18} * 6.02 * 10^{23} = 13.123 * 10^{5} \left(\frac{J}{mol}\right) = 1312.3 \left(\frac{kJ}{mol}\right)$$

Answer: $E = 1312.3 \ \mathrm{kJ} * \mathrm{mol}^{-1}$

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