Calculate wavelength of photon emitted when an electron goes from n=3 to n=2 level of hydrogen atom.

Answer

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

### Step 1: Identify the formula
We will use the formula for the energy of the photon emitted during the transition:
\[ \frac{1}{\lambda} = R \cdot Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
where:
- \( \lambda \) is the wavelength,
- \( R \) is the Rydberg constant (approximately \( 1.097 \times 10^7 \, \text{m}^{-1} \)),
- \( Z \) is the atomic number (for hydrogen, \( Z = 1 \)),
- \( n_1 \) is the lower energy level (2 in this case),
- \( n_2 \) is the higher energy level (3 in this case).

### Step 2: Substitute the values
Substituting the values into the formula:
\[ \frac{1}{\lambda} = R \cdot 1^2 \left( \frac{1}{2^2} - \frac{1}{3^2} \right) \]
Calculating the fractions:
\[ \frac{1}{2^2} = \frac{1}{4}, \quad \frac{1}{3^2} = \frac{1}{9} \]
Thus,
\[ \frac{1}{\lambda} = R \left( \frac{1}{4} - \frac{1}{9} \right) \]

### Step 3: Find a common denominator
To subtract the fractions, we need a common denominator:
\[ \frac{1}{4} = \frac{9}{36}, \quad \frac{1}{9} = \frac{4}{36} \]
So,
\[ \frac{1}{4} - \frac{1}{9} = \frac{9}{36} - \frac{4}{36} = \frac{5}{36} \]

### Step 4: Substitute back into the equation
Now substituting back into the equation:
\[ \frac{1}{\lambda} = R \cdot \frac{5}{36} \]
Using \( R \approx 1.097 \times 10^7 \, \text{m}^{-1} \):
\[ \frac{1}{\lambda} = 1.097 \times 10^7 \cdot \frac{5}{36} \]

### Step 5: Calculate \( \frac{1}{\lambda} \)
Calculating:
\[ \frac{1}{\lambda} \approx 1.097 \times 10^7 \cdot \frac{5}{36} \approx 1.52 \times 10^6 \, \text{m}^{-1} \]

### Step 6: Find \( \lambda \)
Now, take the reciprocal to find \( \lambda \):
\[ \lambda = \frac{1}{1.52 \times 10^6} \approx 6.58 \times 10^{-7} \, \text{m} \]
To convert to nanometers:
\[ \lambda \approx 658 \, \text{nm} \]

### Final Answer
The wavelength of the photon emitted when an electron transitions from n=3 to n=2 level of a hydrogen atom is approximately **658 nm**.