The wavelength of light emitted from second orbit to first orbits in a hydrogen atom is

Options

A. `6563 Å`
B. `4102 Å`
C. `4861 Å`
D. `1215 Å`

Answer

To find the wavelength of light emitted from the second orbit to the first orbit in a hydrogen atom, we can use the Rydberg formula for hydrogen:

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

Where:
- \(\lambda\) is the wavelength of the emitted light,
- \(R\) is the Rydberg constant, approximately \(1.09737 \times 10^7 \, \text{m}^{-1}\),
- \(n_1\) is the principal quantum number of the lower energy level (in this case, \(n_1 = 1\)),
- \(n_2\) is the principal quantum number of the higher energy level (in this case, \(n_2 = 2\)).

### Step 1: Identify the values of \(n_1\) and \(n_2\)
- For the transition from the second orbit to the first orbit, we have:
- \(n_1 = 1\)
- \(n_2 = 2\)

### Step 2: Substitute the values into the Rydberg formula
\[ \frac{1}{\lambda} = R \left( \frac{1}{1^2} - \frac{1}{2^2} \right) \]
Substituting the values:
\[ \frac{1}{\lambda} = 1.09737 \times 10^7 \left( \frac{1}{1} - \frac{1}{4} \right) \]

### Step 3: Simplify the expression
Calculate the term in the parentheses:
\[ \frac{1}{1} - \frac{1}{4} = 1 - 0.25 = 0.75 \]
Now substitute this back into the equation:
\[ \frac{1}{\lambda} = 1.09737 \times 10^7 \times 0.75 \]

### Step 4: Calculate \(\frac{1}{\lambda}\)
\[ \frac{1}{\lambda} = 1.09737 \times 10^7 \times 0.75 = 8.230275 \times 10^6 \, \text{m}^{-1} \]

### Step 5: Calculate \(\lambda\)
Now take the reciprocal to find \(\lambda\):
\[ \lambda = \frac{1}{8.230275 \times 10^6} \approx 1.215 \times 10^{-7} \, \text{m} \]

### Step 6: Convert to Angstroms
To convert meters to Angstroms (1 Angstrom = \(10^{-10}\) m):
\[ \lambda \approx 1.215 \times 10^{-7} \, \text{m} = 1.215 \times 10^{3} \, \text{Angstroms} = 121.5 \, \text{nm} \]

### Final Answer
The wavelength of light emitted from the second orbit to the first orbit in a hydrogen atom is approximately **121.5 nm** or **1215 Angstroms**.

---