The wavelength of the photon emitted by a hydrogen atom when an electron makes a transition from n = 2 to n = 1 state is :

Options

A. 194.8 nm
B. 913.3 nm
C. 490.7 nm
D. 121.8 nm

Answer

To find the wavelength of the photon emitted by a hydrogen atom when an electron transitions from the n = 2 state to the n = 1 state, we can use the Rydberg formula for hydrogen:

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

Where:
- \(\lambda\) is the wavelength of the emitted photon,
- \(R\) is the Rydberg constant (\(1.096 \times 10^7 \, \text{m}^{-1}\)),
- \(Z\) is the atomic number (for hydrogen, \(Z = 1\)),
- \(n_1\) is the lower energy level (1 in this case),
- \(n_2\) is the higher energy level (2 in this case).

### Step 1: Identify the values
For this transition:
- \(n_1 = 1\)
- \(n_2 = 2\)
- \(Z = 1\)

### Step 2: Substitute the values into the Rydberg formula
Substituting the values into the formula:

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

\[ \frac{1}{\lambda} = 1.096 \times 10^7 \left( 1 - \frac{1}{4} \right) \]

### Step 3: Simplify the expression
Calculating the term inside the parentheses:

\[ 1 - \frac{1}{4} = \frac{3}{4} \]

So, we have:

\[ \frac{1}{\lambda} = 1.096 \times 10^7 \cdot \frac{3}{4} \]

### Step 4: Calculate \(\frac{1}{\lambda}\)
Now, calculate \(\frac{1}{\lambda}\):

\[ \frac{1}{\lambda} = 1.096 \times 10^7 \cdot 0.75 = 0.822 \times 10^7 \, \text{m}^{-1} \]

### Step 5: Find \(\lambda\)
Taking the reciprocal to find \(\lambda\):

\[ \lambda = \frac{1}{0.822 \times 10^7} \approx 1.217 \times 10^{-8} \, \text{m} \]

### Step 6: Convert to nanometers
To convert meters to nanometers (1 m = \(10^9\) nm):

\[ \lambda \approx 1.217 \times 10^{-8} \times 10^9 \, \text{nm} = 12.17 \, \text{nm} \]

### Final Answer
Thus, the wavelength of the photon emitted when the electron transitions from n = 2 to n = 1 in a hydrogen atom is approximately:

\[ \lambda \approx 12.17 \, \text{nm} \]