The wavelength of certain line in H-atom spectra is observed to be `4341Å. (R_(H)= 109677cm^(-1))`. The value of quantum number of higher state is:

Options

A. 3
B. 4
C. 5
D. data insufficient

Explanation

To solve the problem, we need to find the value of the quantum number of the higher state (n2) for the transition in the hydrogen atom that corresponds to the given wavelength of 4341 Å. We will use the Rydberg formula for hydrogen spectra.

### Step-by-Step Solution:

1. **Identify the Given Values**:
- Wavelength (λ) = 4341 Å = 4341 x 10^(-8) cm (since 1 Å = 10^(-8) cm)
- Rydberg constant (R_H) = 109677 cm^(-1)
- Lower state quantum number (n1) = 2 (since the wavelength lies in the visible region, which corresponds to the Balmer series).

2. **Use the Rydberg Formula**:
The Rydberg formula for the wavelengths of spectral lines in hydrogen is given by:
\[ \frac{1}{\lambda} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
Substituting the known values into the formula:
\[ \frac{1}{4341 \times 10^{-8}} = 109677 \left( \frac{1}{2^2} - \frac{1}{n_2^2} \right) \]

3. **Calculate \(\frac{1}{\lambda}\)**:
\[ \frac{1}{4341 \times 10^{-8}} = \frac{1}{4.341 \times 10^{-5}} \approx 23040.5 \text{ cm}^{-1} \]

4. **Set Up the Equation**:
Now we can set up the equation:
\[ 23040.5 = 109677 \left( \frac{1}{4} - \frac{1}{n_2^2} \right) \]

5. **Solve for \(\frac{1}{n_2^2}\)**:
First, calculate \(\frac{1}{4}\):
\[ \frac{1}{4} = 0.25 \]
Now substitute this back into the equation:
\[ 23040.5 = 109677 (0.25 - \frac{1}{n_2^2}) \]
Dividing both sides by 109677:
\[ \frac{23040.5}{109677} = 0.25 - \frac{1}{n_2^2} \]
\[ 0.2100 \approx 0.25 - \frac{1}{n_2^2} \]

6. **Rearranging the Equation**:
Rearranging gives:
\[ \frac{1}{n_2^2} = 0.25 - 0.2100 = 0.04 \]

7. **Find \(n_2^2\)**:
Taking the reciprocal:
\[ n_2^2 = \frac{1}{0.04} = 25 \]

8. **Calculate \(n_2\)**:
Taking the square root:
\[ n_2 = 5 \]

### Final Answer:
The value of the quantum number of the higher state (n2) is **5**.