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

Leave a Reply