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. 6

Answer

To solve the problem, we need to find the quantum number of the higher state (n2) for the hydrogen atom given the wavelength of a spectral line. The wavelength is given as \( 4341 \, \text{Å} \) and the Rydberg constant \( R_H = 109677 \, \text{cm}^{-1} \).

### Step 1: Convert Wavelength to Centimeters
First, we convert the wavelength from angstroms to centimeters. Since \( 1 \, \text{Å} = 10^{-10} \, \text{m} \) and \( 1 \, \text{m} = 100 \, \text{cm} \), we can convert:

\[ \lambda = 4341 \, \text{Å} = 4341 \times 10^{-10} \, \text{m} = 4341 \times 10^{-8} \, \text{cm} \]

### Step 2: Use the Rydberg Formula
The Rydberg formula for the wavelength of light emitted during a transition in a hydrogen atom is given by:

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

Where:
- \( \lambda \) is the wavelength,
- \( R_H \) is the Rydberg constant,
- \( n_1 \) is the lower energy level,
- \( n_2 \) is the higher energy level.

### Step 3: Identify Quantum Numbers
In the case of the Balmer series (which includes visible light), the lower state \( n_1 \) is 2. Thus, we have:

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

### Step 4: Substitute Known Values
Substituting the values we have:

\[ \frac{1}{4341 \times 10^{-8}} = 109677 \left( \frac{1}{4} - \frac{1}{n_2^2} \right) \]

Calculating \( \frac{1}{4341 \times 10^{-8}} \):

\[ \frac{1}{4341 \times 10^{-8}} \approx 2.304 \times 10^{7} \, \text{cm}^{-1} \]

### Step 5: Set Up the Equation
Now we set up the equation:

\[ 2.304 \times 10^{7} = 109677 \left( \frac{1}{4} - \frac{1}{n_2^2} \right) \]

### Step 6: Solve for \( n_2^2 \)
Simplifying the equation:

\[ \frac{1}{4} = 0.25 \]

So we can rewrite the equation as:

\[ 2.304 \times 10^{7} = 109677 \left( 0.25 - \frac{1}{n_2^2} \right) \]

Dividing both sides by \( 109677 \):

\[ \frac{2.304 \times 10^{7}}{109677} = 0.25 - \frac{1}{n_2^2} \]

Calculating the left side:

\[ \approx 0.21 = 0.25 - \frac{1}{n_2^2} \]

### Step 7: Rearranging the Equation
Rearranging gives:

\[ \frac{1}{n_2^2} = 0.25 - 0.21 = 0.04 \]

### Step 8: Solve for \( n_2 \)
Taking the reciprocal:

\[ n_2^2 = \frac{1}{0.04} = 25 \]

Taking the square root:

\[ n_2 = 5 \]

### Conclusion
The quantum number of the higher state is \( n_2 = 5 \).

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