If `lamda=c_(2)[(n^(2))/(n^(2)-Z^(2))]` for Balmer series what is the value of `C_(2)`?

Options

A. `r/(R_(H))`
B. `2/(R_(H))`
C. `2R_(H)`
D. `4R_(H)`

Explanation

To find the value of \( C_2 \) in the equation for the Balmer series given by

\[ \lambda = C_2 \frac{n^2}{n^2 - Z^2} \]

we can follow these steps:

### Step 1: Understand the Balmer Series
The Balmer series corresponds to electron transitions in a hydrogen atom where the electron falls to the second energy level (n=2). The formula for the wavelength of emitted light in the Balmer series can be derived from the Rydberg formula:

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

where:
- \( R_H \) is the Rydberg constant for hydrogen,
- \( Z \) is the atomic number (for hydrogen, \( Z = 1 \)),
- \( n_1 \) is the lower energy level (for Balmer series, \( n_1 = 2 \)),
- \( n_2 \) is the higher energy level (where \( n_2 \) can be 3, 4, 5, ...).

### Step 2: Substitute Values into the Rydberg Formula
For the Balmer series, we set \( n_1 = 2 \):

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

This simplifies to:

\[ \frac{1}{\lambda} = R_H \cdot Z^2 \left( \frac{1}{4} - \frac{1}{n_2^2} \right) \]

### Step 3: Rearranging the Equation
Rearranging gives:

\[ \lambda = \frac{1}{R_H \cdot Z^2 \left( \frac{1}{4} - \frac{1}{n_2^2} \right)} \]

### Step 4: Find a Common Denominator
To combine the terms in the parentheses:

\[ \frac{1}{4} - \frac{1}{n_2^2} = \frac{n_2^2 - 4}{4n_2^2} \]

Thus, we have:

\[ \lambda = \frac{4n_2^2}{R_H \cdot Z^2 (n_2^2 - 4)} \]

### Step 5: Compare with Given Formula
Now, we compare this with the given formula:

\[ \lambda = C_2 \frac{n^2}{n^2 - Z^2} \]

From our derived equation, we see that:

\[ C_2 = \frac{4}{R_H \cdot Z^2} \]

### Step 6: Substitute \( Z \)
For hydrogen, \( Z = 1 \):

\[ C_2 = \frac{4}{R_H \cdot 1^2} = \frac{4}{R_H} \]

### Step 7: Final Value of \( C_2 \)
Thus, the value of \( C_2 \) is:

\[ C_2 = \frac{4}{R_H} \]

### Conclusion
The final answer is:

\[ C_2 = \frac{4}{R_H} \]