Calculate the wavelength and energy for radiation emitted for the electron transition from infinite `(oo)` to stationary state of the hydrogen atom
`R = 1.0967 xx 10^(7) m^(-1), h = 6.6256 xx 10^(-34) J s ` and `c = 2.979 xx 10^(8) m s^(-1)`

Explanation

To solve the problem of calculating the wavelength and energy for radiation emitted during the electron transition from infinite to the stationary state of the hydrogen atom, we can follow these steps:

### Step 1: Understand the Transition
The transition is from a higher energy level (n = ∞) to the ground state (n = 1) of the hydrogen atom. The Rydberg formula for the wavelength of emitted radiation is given by:

\[ \frac{1}{\lambda} = R \cdot Z^2 \left( \frac{1}{n_l^2} - \frac{1}{n_h^2} \right) \]

Where:
- \( R \) is the Rydberg constant,
- \( Z \) is the atomic number (for hydrogen, \( Z = 1 \)),
- \( n_l \) is the lower energy level (1 in this case),
- \( n_h \) is the higher energy level (∞ in this case).

### Step 2: Substitute Values into the Rydberg Formula
Given:
- \( R = 1.0967 \times 10^7 \, \text{m}^{-1} \)
- \( Z = 1 \)
- \( n_l = 1 \)
- \( n_h = \infty \)

Substituting these values into the Rydberg formula:

\[ \frac{1}{\lambda} = 1.0967 \times 10^7 \cdot 1^2 \left( \frac{1}{1^2} - \frac{1}{\infty^2} \right) \]

Since \( \frac{1}{\infty^2} = 0 \), the equation simplifies to:

\[ \frac{1}{\lambda} = 1.0967 \times 10^7 \cdot (1 - 0) = 1.0967 \times 10^7 \]

### Step 3: Calculate Wavelength
Now, we can find the wavelength \( \lambda \):

\[ \lambda = \frac{1}{1.0967 \times 10^7} \approx 9.11 \times 10^{-8} \, \text{m} \]

### Step 4: Calculate Energy
To find the energy \( \Delta E \) of the emitted radiation, we use the formula:

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

Where:
- \( h = 6.6256 \times 10^{-34} \, \text{J s} \) (Planck's constant),
- \( c = 2.979 \times 10^8 \, \text{m/s} \) (speed of light).

Substituting the values:

\[ \Delta E = \frac{(6.6256 \times 10^{-34}) \cdot (2.979 \times 10^8)}{9.11 \times 10^{-8}} \]

Calculating this gives:

\[ \Delta E \approx 2.166 \times 10^{-18} \, \text{J} \]

Rounding this to two decimal places gives:

\[ \Delta E \approx 2.17 \times 10^{-18} \, \text{J} \]

### Final Results
- Wavelength \( \lambda \approx 9.11 \times 10^{-8} \, \text{m} \)
- Energy \( \Delta E \approx 2.17 \times 10^{-18} \, \text{J} \)