Given the value of Rydberg constant is `10^(7)m^(-1)`, the waves number of the lest line of the Balmer series in hydrogen spectrum will be:

Options

A. `0.025 xx 10^(4)m^(-1)`
B. `0.5 xx 10^(7)m^(-1)`
C. `0.25 xx 10^(7)m^(-1)`
D. `2.5 xx 10^(7)m^(-1)`

Explanation

To find the wave number of the last line of the Balmer series in the hydrogen spectrum, we can follow these steps:

### Step 1: Understand the Balmer Series
The Balmer series corresponds to transitions where an electron falls to the second energy level (n=2) from higher energy levels (n=3, 4, 5, ...). The last line of the Balmer series occurs when the electron transitions from infinity (n=∞) to n=2.

### Step 2: Identify the Formula for Wave Number
The wave number (ṽ) can be calculated using the Rydberg formula:
\[ \tilde{\nu} = RZ^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
where:
- \( R \) is the Rydberg constant,
- \( Z \) is the atomic number (for hydrogen, \( Z = 1 \)),
- \( n_1 \) is the lower energy level (for the Balmer series, \( n_1 = 2 \)),
- \( n_2 \) is the higher energy level (for the last line of the Balmer series, \( n_2 = \infty \)).

### Step 3: Substitute Values into the Formula
We know:
- \( R = 10^7 \, \text{m}^{-1} \)
- \( Z = 1 \)
- \( n_1 = 2 \)
- \( n_2 = \infty \)

Substituting these values into the formula:
\[ \tilde{\nu} = 10^7 \cdot 1^2 \left( \frac{1}{2^2} - \frac{1}{\infty^2} \right) \]

### Step 4: Simplify the Expression
Since \( \frac{1}{\infty^2} = 0 \), we can simplify:
\[ \tilde{\nu} = 10^7 \left( \frac{1}{4} - 0 \right) = 10^7 \cdot \frac{1}{4} \]

### Step 5: Calculate the Wave Number
Calculating the wave number gives:
\[ \tilde{\nu} = \frac{10^7}{4} = 2.5 \times 10^6 \, \text{m}^{-1} \]

### Step 6: Final Result
Thus, the wave number of the last line of the Balmer series in the hydrogen spectrum is:
\[ \tilde{\nu} = 2.5 \times 10^6 \, \text{m}^{-1} \]