Calculate the wavelength of radiation emited when an electron in a hydrogen atom makes a transition from an energy level with `n = 3` to a level with `n= 2`

Explanation

To calculate the wavelength of radiation emitted when an electron in a hydrogen atom transitions from an energy level with \( n = 3 \) to \( n = 2 \), we can follow these steps:

### Step 1: Identify the Transition Levels
We have the initial energy level \( n_2 = 3 \) and the final energy level \( n_1 = 2 \).

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

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

where:
- \( \lambda \) is the wavelength,
- \( R_H \) is the Rydberg constant for hydrogen, approximately \( 109678 \, \text{cm}^{-1} \),
- \( n_1 \) is the lower energy level (2 in this case),
- \( n_2 \) is the higher energy level (3 in this case).

### Step 3: Substitute the Values into the Formula
Substituting \( n_1 = 2 \) and \( n_2 = 3 \) into the formula:

\[ \frac{1}{\lambda} = 109678 \left( \frac{1}{2^2} - \frac{1}{3^2} \right) \]

### Step 4: Calculate the Terms Inside the Parentheses
Calculating the squares:

\[ \frac{1}{2^2} = \frac{1}{4} = 0.25 \]
\[ \frac{1}{3^2} = \frac{1}{9} \approx 0.1111 \]

Now, substituting these values:

\[ \frac{1}{\lambda} = 109678 \left( 0.25 - 0.1111 \right) \]
\[ \frac{1}{\lambda} = 109678 \left( 0.1389 \right) \]

### Step 5: Calculate \( \frac{1}{\lambda} \)
Now, calculating:

\[ \frac{1}{\lambda} = 109678 \times 0.1389 \approx 15238.6 \, \text{cm}^{-1} \]

### Step 6: Calculate \( \lambda \)
To find \( \lambda \), take the reciprocal:

\[ \lambda = \frac{1}{15238.6} \approx 6.56 \times 10^{-5} \, \text{cm} \]

### Step 7: Convert to Nanometers
To convert centimeters to nanometers (1 cm = \( 10^7 \) nm):

\[ \lambda \approx 6.56 \times 10^{-5} \, \text{cm} \times 10^7 \, \text{nm/cm} \approx 656 \, \text{nm} \]

### Final Answer
The wavelength of the emitted radiation is approximately **656 nm**.

---