The wavelength of the radiation emitted , when in a hydrogen atom electron falls from infinity to stationary state 1 , would be :
(Rydberg constant = `1.097 xx 10^(7) m^(-1)`)

Options

A. 192 nm
B. `4.6nm`
C. `9.1xx10^(-8)` nm
D. 91 nm

Answer

To solve the problem of finding the wavelength of the radiation emitted when an electron in a hydrogen atom falls from infinity to the stationary state 1, we can use the Rydberg formula for hydrogen. The formula is given by:

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

Where:
- \(\lambda\) is the wavelength of the emitted radiation,
- \(R\) is the Rydberg constant (\(1.097 \times 10^7 \, \text{m}^{-1}\)),
- \(n_1\) is the principal quantum number of the lower energy level,
- \(n_2\) is the principal quantum number of the higher energy level.

### Step 1: Identify the values of \(n_1\) and \(n_2\)
In this case:
- The electron falls from infinity (\(n_2 = \infty\)) to the stationary state 1 (\(n_1 = 1\)).

### Step 2: Substitute the values into the Rydberg formula
Now, substituting \(n_1 = 1\) and \(n_2 = \infty\) into the Rydberg formula:

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

Since \(\frac{1}{\infty^2} = 0\), we can simplify this to:

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

### Step 3: Substitute the value of the Rydberg constant
Now, substituting the value of the Rydberg constant:

\[ \frac{1}{\lambda} = 1.097 \times 10^7 \, \text{m}^{-1} \]

### Step 4: Calculate \(\lambda\)
To find \(\lambda\), we take the reciprocal:

\[ \lambda = \frac{1}{1.097 \times 10^7} \, \text{m} \]

Calculating this gives:

\[ \lambda \approx 9.1 \times 10^{-8} \, \text{m} \]

### Step 5: Convert to nanometers
To convert meters to nanometers (1 m = \(10^9\) nm):

\[ \lambda \approx 9.1 \times 10^{-8} \, \text{m} = 91 \, \text{nm} \]

### Final Answer
The wavelength of the radiation emitted when the electron falls from infinity to stationary state 1 is approximately **91 nm**.