The frequency of series limit of Balmer series of hydrogen atom in terms of Rydberg constant R and velocity of light (C ) is

Options

A. 2RC
B. `(RC)/(2)`
C. `(RC)/(4)`
D. 4RC

Explanation

To find the frequency of the series limit of the Balmer series of the hydrogen atom in terms of the Rydberg constant \( R \) and the speed of light \( c \), we can follow these steps:

### Step 1: Understand the Balmer Series
The Balmer series corresponds to electronic transitions in a hydrogen atom where the final energy level is \( n = 2 \). The general formula for the wavelengths of the spectral lines in the hydrogen atom is given by:

\[ \frac{1}{\lambda} = R \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right) \]

where:
- \( R \) is the Rydberg constant,
- \( n_f \) is the final energy level,
- \( n_i \) is the initial energy level.

### Step 2: Identify the Series Limit
For the series limit of the Balmer series, the electron transitions from an infinitely high energy level (\( n_i \to \infty \)) to \( n_f = 2 \). Therefore, we can substitute these values into the formula:

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

Since \( \frac{1}{\infty^2} = 0 \), the equation simplifies to:

\[ \frac{1}{\lambda} = R \left( \frac{1}{4} \right) \]

### Step 3: Calculate Wavelength
Rearranging the equation gives us the wavelength \( \lambda \):

\[ \lambda = \frac{4}{R} \]

### Step 4: Relate Wavelength to Frequency
We know that the speed of light \( c \) is related to frequency \( f \) and wavelength \( \lambda \) by the equation:

\[ c = f \lambda \]

From this, we can express frequency \( f \) as:

\[ f = \frac{c}{\lambda} \]

### Step 5: Substitute for Wavelength
Now, substitute the expression for \( \lambda \) into the frequency equation:

\[ f = \frac{c}{\frac{4}{R}} = \frac{cR}{4} \]

### Final Result
Thus, the frequency of the series limit of the Balmer series of the hydrogen atom in terms of the Rydberg constant \( R \) and the speed of light \( c \) is:

\[ f = \frac{cR}{4} \]

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