According to bohr's theory, the wave number of last line of balmer series is `(R =1.1xx10^(7)m^(-1))`

Options

A. `5.5xx10^5m^(-1)`
B. `4.4xx10^7m^(-1)`
C. `2.75xx10^6m^(-1)`
D. `2.75xx10^8m^(-1)`

Explanation

To find the wave number of the last line of the Balmer series according to Bohr's theory, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Balmer Series**: The Balmer series corresponds to the transitions of electrons in a hydrogen atom from higher energy levels (n ≥ 3) to the second energy level (n1 = 2).

2. **Identify the Last Line of the Balmer Series**: The last line of the Balmer series occurs when the electron transitions from the highest energy level (n2 = ∞) to n1 = 2.

3. **Use the Formula for Wave Number**: The wave number (σ) is given by the formula:
\[ \sigma = \frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
where \( R \) is the Rydberg constant and \( n_1 \) and \( n_2 \) are the principal quantum numbers.

4. **Substitute Values**: For the last line of the Balmer series:
- \( n_1 = 2 \)
- \( n_2 = \infty \)

Therefore, the equation becomes:
\[ \sigma = R \left( \frac{1}{2^2} - \frac{1}{\infty^2} \right) \]

5. **Calculate the Terms**: Since \( \frac{1}{\infty^2} = 0 \), the equation simplifies to:
\[ \sigma = R \left( \frac{1}{4} - 0 \right) = \frac{R}{4} \]

6. **Substitute the Value of R**: Given that \( R = 1.1 \times 10^7 \, m^{-1} \):
\[ \sigma = \frac{1.1 \times 10^7}{4} \]

7. **Perform the Calculation**:
\[ \sigma = 2.75 \times 10^6 \, m^{-1} \]

### Final Answer:
The wave number of the last line of the Balmer series is \( 2.75 \times 10^6 \, m^{-1} \).

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