Calculate the wavelength and energy of radiation emitted for the electronic transition from infinity `(oo)` to stationary state one of the hydrogen atom `("R = 1.09678 "xx 10^(7) m^(-1))`

Explanation

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

### Step 1: Identify the transition states
In this case, the transition is from infinity (n2 = ∞) to the first stationary state (n1 = 1) of the hydrogen atom.

### Step 2: Use the Rydberg formula for wavelength
The Rydberg formula for the wavelength of emitted radiation is given by:

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

Where:
- \(\lambda\) is the wavelength,
- \(R\) is the Rydberg constant (\(1.09678 \times 10^7 \, \text{m}^{-1}\)),
- \(n_1\) is the final state (1),
- \(n_2\) is the initial state (∞).

### Step 3: Substitute the values into the formula
Substituting \(n_1 = 1\) and \(n_2 = \infty\):

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

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

\[ \frac{1}{\lambda} = R \left( 1 - 0 \right) = R \]

Thus,

\[ \lambda = \frac{1}{R} \]

### Step 4: Calculate the wavelength
Now, substituting the value of \(R\):

\[ \lambda = \frac{1}{1.09678 \times 10^7} \approx 9.11 \times 10^{-8} \, \text{m} \]

### Step 5: Calculate the energy of the emitted radiation
The energy of the emitted radiation can be calculated using the formula:

\[ E = h \nu \]

Where:
- \(E\) is the energy,
- \(h\) is Planck's constant (\(6.626 \times 10^{-34} \, \text{J s}\)),
- \(\nu\) is the frequency.

The frequency \(\nu\) can be calculated using the speed of light \(c\) and the wavelength \(\lambda\):

\[ \nu = \frac{c}{\lambda} \]

Substituting the values:

\[ c = 3 \times 10^8 \, \text{m/s} \]

\[ \nu = \frac{3 \times 10^8}{9.11 \times 10^{-8}} \approx 3.29 \times 10^{15} \, \text{Hz} \]

### Step 6: Calculate the energy
Now substituting \(\nu\) into the energy formula:

\[ E = h \nu = (6.626 \times 10^{-34}) \times (3.29 \times 10^{15}) \approx 2.17 \times 10^{-18} \, \text{J} \]

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

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