Calculate the ionisation energy of H atom.

Explanation

To calculate the ionization energy of a hydrogen atom, we will follow these steps:

### Step 1: Understand the Concept of Ionization Energy
Ionization energy is the energy required to remove an electron from an atom. For hydrogen, we are interested in removing the electron from its ground state (n=1) to an infinite distance (n=∞).

### Step 2: Use Bohr's Model of the Atom
According to Bohr's model, the relationship between the wavelengths of emitted or absorbed light and the energy levels of an atom can be expressed as:

\[ \frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]

Where:
- \( \lambda \) is the wavelength of the emitted or absorbed light,
- \( R \) is the Rydberg constant (\( R = 1.0974 \times 10^7 \, \text{m}^{-1} \)),
- \( n_1 \) is the lower energy level (for ionization, \( n_1 = 1 \)),
- \( n_2 \) is the higher energy level (for ionization, \( n_2 = \infty \)).

### Step 3: Substitute Values into the Equation
Substituting the values into the equation:

\[ \frac{1}{\lambda} = 1.0974 \times 10^7 \left( \frac{1}{1^2} - \frac{1}{\infty^2} \right) \]

Since \( \frac{1}{\infty^2} = 0 \):

\[ \frac{1}{\lambda} = 1.0974 \times 10^7 \left( 1 - 0 \right) = 1.0974 \times 10^7 \]

### Step 4: Calculate Wavelength \( \lambda \)
Now, we can find \( \lambda \):

\[ \lambda = \frac{1}{1.0974 \times 10^7} \approx 9.116 \times 10^{-8} \, \text{m} \]

### Step 5: Calculate the Energy Using the Wavelength
The energy \( E \) can be calculated using the formula:

\[ E = \frac{hc}{\lambda} \]

Where:
- \( h \) is Planck's constant (\( h = 6.63 \times 10^{-34} \, \text{J s} \)),
- \( c \) is the speed of light (\( c = 3 \times 10^8 \, \text{m/s} \)).

Substituting the values:

\[ E = \frac{(6.63 \times 10^{-34})(3 \times 10^8)}{9.116 \times 10^{-8}} \]

### Step 6: Perform the Calculation
Calculating the above expression:

\[ E \approx \frac{1.989 \times 10^{-25}}{9.116 \times 10^{-8}} \approx 2.181 \times 10^{-18} \, \text{J} \]

### Conclusion
The ionization energy of the hydrogen atom is approximately:

\[ E \approx 2.181 \times 10^{-18} \, \text{J} \]