Question

How do you calculate the ionization energy of a hydrogen atom in its ground state?

Answer 1

Step 1: Rydberg expression for calculating Ionization energy:-

The amount of energy required to extract an electron from an isolated gaseous atom to form a positive ion is called Ionization energy.

For calculating the Ionization energy of a hydrogen atom in its ground state we can use the Rydberg expression given as:

$\frac{1}{\mathrm{\lambda }}=\mathrm{R}\left[\frac{1}{{{\mathrm{n}}_1}^2}-\frac{1}{{{\mathrm{n}}_2}^2}\right]$

( Where $\mathrm{\lambda }=$Wavelength of the electron

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

${\mathrm{n}}_1$= the principal quantum number of the lower energy level,

${\mathrm{n}}_2$ = the principal quantum number of the higher energy level.)

Step 2: Calculate the value of ${\mathrm{n}}_1$ and ${\mathrm{n}}_2$:-

As the electron exists in the ground${\mathrm{n}}_1=1$ground state we have considered series 1 from which the transitions occur is known as The Lyman Series.

Now, we know that as the value of ${\mathrm{n}}_2$ increases, the value of $\frac{1}{{\mathrm{n}}_2}$decreases. When$\mathrm{n}=\infty$, you can say that $\frac{1}{{\mathrm{n}}_2}\to 0$

Step 3: Calculating the value of $\mathrm{\lambda }$ from the Rydberg expression:-

To calculate the ionization energy of the hydrogen atom in its ground state the Rydberg expression can be written as;

$$\begin{gathered} \frac{1}{\mathrm{\lambda }}=\mathrm{R}\left[\frac{1}{{{\mathrm{n}}_1}^2}-0\right] \\ =\mathrm{R}\times \frac{1}{{{\mathrm{n}}_1}^2} \end{gathered}$$

$\because$ ${\mathrm{n}}_1=1$, so the expression becomes:$\frac{1}{\mathrm{\lambda }}=R$

Step 4: Calculating frequency:-

$$\begin{gathered} \Rightarrow \frac{1}{\mathrm{\lambda }}=1.097\times {10}^7 \\ \Rightarrow \mathrm{\lambda }=9.116\times {10}^{-8}\mathrm{m} \\ \because \mathrm{c}=\mathrm{\nu \lambda } \\ \Rightarrow \mathrm{\nu }=\frac{\mathrm{c}}{\mathrm{\lambda }} \end{gathered}$$

Here, $\mathrm{C}=$the velocity of light $=3\times {10}^8\mathrm{m}/\mathrm{s}$

and $\mathrm{v}$= Frequency $\left({\mathrm{s}}^{-1}\right)$

Hence,

$$\begin{gathered} \mathrm{v}=\frac{3\times {10}^8}{9.116\times {10}^{-8}} \\ =3.291\times {10}^{15}{\mathrm{s}}^{-1} \end{gathered}$$

Step 5: Calculating ionization energy:-

We know, $\mathrm{E}=\mathrm{h\nu }$

(Where$\mathrm{E}=$ The ionization energy of the Hydrogen atom

$\mathrm{h}=$ planks constant $=6.626\times {10}^{-34}\mathrm{J}/\mathrm{s}$

$\mathrm{v}$= Frequency )

Hence, the ionization energy can be calculated as:

$$\begin{gathered} \Rightarrow \mathrm{E}=6.626\times {10}^{-34}\times 3.291\times {10}^{15} \\ =2.18\times {10}^{-18}\mathrm{J} \\ =2.18\times {10}^{-18}\times 6.02\times {10}^{23} \\ =13.123\times {10}^5\mathrm{J}{\mathrm{mol}}^{-1} \\ =1312\mathrm{kJ}{\mathrm{mol}}^{-1} \end{gathered}$$

Hence, Using the Rydberg expression, the Ionization energy of a hydrogen atom in its ground state can be calculated to be $1312\mathrm{kJ}{\mathrm{mol}}^{-1}.$