The wavelength of the ultraviolet region of the hydrogen spectrum is 122 nm. The wavelength of the second sperctral line in the Balmer series of singly ionized helium atom is (a) `1215 Å` (b) `1640 Å` (C ) `2430 Å` (d) `4687 Å`

Explanation

To solve the problem of finding the wavelength of the second spectral line in the Balmer series of singly ionized helium, we will follow these steps:

### Step 1: Understand the Given Information
We know that the wavelength of the ultraviolet region of the hydrogen spectrum is given as 122 nm. This corresponds to the Lyman series transition from n=2 to n=1.

### Step 2: Use the Rydberg Formula
The Rydberg formula for the wavelengths of spectral lines in hydrogen-like atoms is given by:

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

Where:
- \( R \) is the Rydberg constant (approximately \( 1.097 \times 10^7 \, \text{m}^{-1} \))
- \( Z \) is the atomic number
- \( n_f \) is the final energy level
- \( n_i \) is the initial energy level

### Step 3: Apply the Formula for Hydrogen
For hydrogen (Z = 1), the transition from n=2 to n=1 gives us:

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

Calculating this gives:

\[ \frac{1}{\lambda} = R \left( 1 - \frac{1}{4} \right) = R \cdot \frac{3}{4} \]

### Step 4: Apply the Formula for Helium
For singly ionized helium (Z = 2), we want to find the second spectral line in the Balmer series, which corresponds to a transition from n=4 to n=2 (the second line corresponds to n=3 to n=2, but we are looking for the second line overall).

Using the Rydberg formula for helium:

\[ \frac{1}{\lambda'} = R \cdot 2^2 \left( \frac{1}{2^2} - \frac{1}{4^2} \right) \]

Calculating this gives:

\[ \frac{1}{\lambda'} = R \cdot 4 \left( \frac{1}{4} - \frac{1}{16} \right) = R \cdot 4 \left( \frac{4 - 1}{16} \right) = R \cdot 4 \cdot \frac{3}{16} = R \cdot \frac{3}{4} \]

### Step 5: Compare the Two Equations
From the calculations, we see that:

\[ \frac{1}{\lambda} = R \cdot \frac{3}{4} \quad \text{(for hydrogen)} \]
\[ \frac{1}{\lambda'} = R \cdot \frac{3}{4} \quad \text{(for helium)} \]

This means:

\[ \lambda' = \lambda \]

### Step 6: Convert Wavelength to Appropriate Units
We know that \( \lambda = 122 \, \text{nm} = 1220 \, \text{Å} \).

### Conclusion
Thus, the wavelength of the second spectral line in the Balmer series of singly ionized helium is also 122 nm or 1220 Å. However, since the options given in the question do not include this value, it indicates that there might be an error in the options provided.

### Final Answer
None of the options are correct.