What would be the wavelength and name of series respectively for the emission transition for H-atom if it starts from the orbit having radius 1.3225 nm and ends at 211.6 pm?

Options

A. 434 nm, Balmer
B. 434 pm, Paschen
C. 545 pm, Pfund
D. 600 nm, Lyman

Explanation

To solve the problem, we need to determine the wavelength and the name of the series for the emission transition of a hydrogen atom, starting from a radius of 1.3225 nm and ending at 211.6 pm.

### Step-by-Step Solution:

1. **Convert Units**:
- Convert the given radii into the same unit. We have:
- \( R_1 = 1.3225 \, \text{nm} = 1.3225 \times 10^{-9} \, \text{m} = 13225 \, \text{pm} \)
- \( R_2 = 211.6 \, \text{pm} \)

2. **Use the Formula for Radius of Orbit**:
- The radius of the nth orbit for hydrogen is given by:
\[ R_n = \frac{0.529 \, \text{Å} \cdot n^2}{Z} \]
- Since \( 1 \, \text{Å} = 100 \, \text{pm} \), we can convert this to picometers:
\[ R_n = \frac{52.9 \, \text{pm} \cdot n^2}{Z} \]
- For hydrogen, \( Z = 1 \).

3. **Set Up the Equations**:
- For \( R_1 \):
\[ R_1 = 52.9 \cdot n_1^2 \implies 13225 = 52.9 \cdot n_1^2 \]
- For \( R_2 \):
\[ R_2 = 52.9 \cdot n_2^2 \implies 211.6 = 52.9 \cdot n_2^2 \]

4. **Calculate \( n_1 \) and \( n_2 \)**:
- From the first equation:
\[ n_1^2 = \frac{13225}{52.9} \implies n_1^2 \approx 250.47 \implies n_1 \approx 15.8 \quad (\text{round to } 16) \]
- From the second equation:
\[ n_2^2 = \frac{211.6}{52.9} \implies n_2^2 \approx 4 \implies n_2 = 2 \]

5. **Determine the Transition**:
- The transition is from \( n_1 = 16 \) to \( n_2 = 2 \).
- The series is determined by the lower energy level. Since the transition ends at \( n_2 = 2 \), it corresponds to the **Balmer series**.

6. **Calculate the Wavelength**:
- Using the Rydberg formula for wavelength:
\[ \frac{1}{\lambda} = R \left( \frac{1}{n_2^2} - \frac{1}{n_1^2} \right) \]
- Where \( R \) (Rydberg constant) is approximately \( 1.097 \times 10^7 \, \text{m}^{-1} \).
- Plugging in the values:
\[ \frac{1}{\lambda} = 1.097 \times 10^7 \left( \frac{1}{2^2} - \frac{1}{16^2} \right) \]
\[ = 1.097 \times 10^7 \left( \frac{1}{4} - \frac{1}{256} \right) \]
\[ = 1.097 \times 10^7 \left( \frac{64 - 1}{256} \right) = 1.097 \times 10^7 \left( \frac{63}{256} \right) \]
\[ = 1.097 \times 10^7 \times 0.2461 \approx 2.70 \times 10^6 \, \text{m}^{-1} \]
- Therefore, the wavelength \( \lambda \) is:
\[ \lambda = \frac{1}{2.70 \times 10^6} \approx 370 \, \text{nm} \]

### Final Answer:
- The wavelength is approximately **370 nm** and the series is the **Balmer series**.