The Ionization energy, radius and velocity of a Hydrogen atom | Quantum Science Philippines

## The Ionization energy, radius and velocity of a Hydrogen atom

a. Show that $E_I= 13.6 eV$

The Ionization energy is the energy needed to remove the electron from an atom. The formula for the Ionization energy of a hydrogen atom in ground state is given by,

$E_I = \frac {{\mu}{e}^{4}}{2 \hbar}$

First we must obtain the values of $\mu$ and $e$ .

$\mu$ is the reduced mass with the formula:

$\mu = \frac{m_em_p}{m_e+m_p}$

where,

$m_e= 0.91 \times 10^{-30} kg$ (mass of the electron) and

$m_p= 1.7 \times 10^{-27} kg$ (mass of the proton).

Directly substituing these values we obtain

$\mu = \frac{0.91 \times 10^{-30} \cdot 1.7 \times 10^{-27}}{0.91 \times 10^{-30} + 1.7 \times 10^{-27}} = 9.1 \times 10^{-31} kg$

The interaction between the proton and electron is electrostatic and will thus have the corresponding potential energy:

$V(r) = \text{-} \frac{q^2}{4\pi {\epsilon}_0}\frac{1}{r} = \text{-} \frac{e^2}{r}$

in which  $r$ denotes the distance between the two particles and e is

$e^2 = \frac{q^2}{4\pi {\epsilon}_0}$

where the electron charge  $q= 1.6 \times 10^{-12} C$

and permittivity of free space ${\epsilon}_0= 8.85 \times 10^{-12} \frac{ C^2}{N{m}^{2}}$

Inputting the above values, we get

$e^2 = \frac{ \big({1.6 \times 10^{-12} C} \big)^2}{4\pi \big( 8.85 \times 10^{-12} \frac{ C^2}{N{m}^{2}}\big)} = 2.32 \times 10^{-28} N{m}^{2}$

Now, substituting the obtained $\mu$ and $e^2$ into $E_I$ ,

$E_I= \frac{{\mu}{e}^{4}}{2 \hbar}$ $\text {where the Planck's constant is}$ $\hbar= 1.055 \times 10^{-34} Js$

$E_I= \frac{{ \big(0.91 \times 10^{-30} kg\big)}{\big(2.32 \times 10^{-28} N{m}^{2} }\big)^{2}}{2 \big(1.055 \times 10^{-34} Js \big)^2}$

$E_I =2.183 \times 10^{-18} J$

Converting this to eV,

$E_I= 2.183 \times 10^{-18} J \cdot \frac{6.242 \times 10^{-18} eV}{1 J}$

Thus, the ionization energy for hydrogen atom is,

$E_I =13.6 eV$

b. Show that  $a_0= 0.53 \AA$

The Bohr radius $a_0$ is the parameter that characterized atomic dimensions.

$a_0= \frac{{\hbar}^2}{\mu {e}^2}$

Substituting the values from part (a) we have,

$\begin{array}{ccc} a_0 &=& \frac{\big({ 1.055 \times 10^{-34} Js}\big)^2}{\big( 0.91 \times 10^{-30} kg\big)\big(2.32 \times 10^{-28} Nm\big)} \\ &=& 5.3 \times 10^{-11} m \end{array}$

Converting to Armstrong, we finally show that

$a_0= 0.53 \AA$

c. Solve for $v_0$

$v_0= \frac{e^2}{\hbar}$

$v_0= \frac{\big(2.32 \times 10^{-28} Nm\big)^2}{1.055 \times 10^{-34} Js}$

$v_0= 2.2 \times 10^6 m/s$