The wavelength of the spectral line, For the transition n = 2 to n=1 in the hydrogen emission spectrum is 'X' . Which of the following spectral line also has a wavelength of X

Options

A. `'H_(alpha)'` line in the Balmer series of spectrum of 'H'
B. `'H_(beta)'` line in the Balmer series of spectrum of 'H'
C. `'H_(alpha)'` line in the Balmer series of spectrum of `'He^(+)'`
D. `'H_(beta)'` line in the Balmer series of spectrum of `'He^(+)'`

Answer

To solve the question regarding the wavelength of the spectral line for the transition from n = 2 to n = 1 in the hydrogen emission spectrum, we will follow these steps:

### Step 1: Understand the Transition
The transition from n = 2 to n = 1 in the hydrogen atom corresponds to the emission of a photon. The wavelength of this emitted photon can be calculated using the Rydberg formula.

### Step 2: Apply the Rydberg Formula
The Rydberg formula is given by:

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

Where:
- \( R_H \) is the Rydberg constant for hydrogen (approximately \( 1.097 \times 10^7 \, \text{m}^{-1} \))
- \( n_1 \) is the lower energy level (1 in this case)
- \( n_2 \) is the higher energy level (2 in this case)

### Step 3: Calculate \( \frac{1}{\lambda} \)
Substituting \( n_1 = 1 \) and \( n_2 = 2 \):

\[ \frac{1}{\lambda} = R_H \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = R_H \left( 1 - \frac{1}{4} \right) = R_H \left( \frac{3}{4} \right) \]

### Step 4: Find the Wavelength \( \lambda \)
Rearranging the equation gives:

\[ \lambda = \frac{4}{3 R_H} \]

### Step 5: Identify Equivalent Transitions
Now, we need to determine which of the provided transitions (H-alpha, H-beta, etc.) also results in the same wavelength \( \lambda \).

1. **H-alpha (n=3 to n=2)**:
\[ \frac{1}{\lambda} = R_H \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = R_H \left( \frac{1}{4} - \frac{1}{9} \right) = R_H \left( \frac{5}{36} \right) \]

2. **H-beta (n=4 to n=2)**:
\[ \frac{1}{\lambda} = R_H \left( \frac{1}{2^2} - \frac{1}{4^2} \right) = R_H \left( \frac{1}{4} - \frac{1}{16} \right) = R_H \left( \frac{3}{16} \right) \]

3. **He+ (n=3 to n=2)**:
\[ \frac{1}{\lambda} = 4R_H \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = 4R_H \left( \frac{5}{36} \right) \]

4. **He+ (n=4 to n=2)**:
\[ \frac{1}{\lambda} = 4R_H \left( \frac{1}{2^2} - \frac{1}{4^2} \right) = 4R_H \left( \frac{3}{16} \right) \]

### Step 6: Compare Values
We need to check if any of these values match \( \frac{3R_H}{4} \).

- The calculations show that none of the transitions from H-alpha or H-beta match \( \frac{4}{3R_H} \).
- However, the transition from He+ (n=4 to n=2) will yield the same wavelength due to the factor of \( z^2 \) (where \( z = 2 \) for He+).

### Conclusion
The spectral line that also has a wavelength of \( X \) is the transition from n=4 to n=2 in the He+ ion.