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

Answer

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 of electrons in a hydrogen atom that end at the energy level \( n_f = 2 \). The initial energy level for the last line of the Balmer series is \( n_i = \infty \).

### Step 2: Use the Rydberg Formula
The wave number \( \bar{\nu} \) (in \( m^{-1} \)) can be calculated using the Rydberg formula:
\[ \bar{\nu} = R \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right) \]
where \( R \) is the Rydberg constant.

### Step 3: Substitute the Values
Given:
- \( R = 10^7 \, m^{-1} \)
- \( n_f = 2 \)
- \( n_i = \infty \)

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

### Step 4: Simplify the Expression
Since \( \frac{1}{\infty^2} = 0 \), the equation simplifies to:
\[ \bar{\nu} = 10^7 \left( \frac{1}{4} - 0 \right) = 10^7 \cdot \frac{1}{4} \]

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

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