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

Options

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

Answer

To solve the question, we need to find the value of \( c_2 \) in the given equation for the Balmer series:

\[ \lambda = c_2 \left( \frac{n^2}{n^2 - 2^2} \right) \]

### Step-by-Step Solution:

1. **Understand the Balmer Series**:
The Balmer series corresponds to transitions in a hydrogen atom where the electron falls to the second energy level (n=2) from higher energy levels (n=3, 4, 5,...).

2. **Use the Formula for Wavelength**:
The formula for the wavelength of the emitted light in the Balmer series can be derived from the Rydberg formula:
\[ \frac{1}{\lambda} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
where \( R_H \) is the Rydberg constant, \( n_1 = 2 \) (for Balmer series), and \( n_2 = n \) (where \( n \) is greater than 2).

3. **Substituting Values**:
Substitute \( n_1 = 2 \) into the formula:
\[ \frac{1}{\lambda} = R_H \left( \frac{1}{2^2} - \frac{1}{n^2} \right) = R_H \left( \frac{1}{4} - \frac{1}{n^2} \right) \]

4. **Finding a Common Denominator**:
The expression can be rewritten as:
\[ \frac{1}{\lambda} = R_H \left( \frac{n^2 - 4}{4n^2} \right) \]
Therefore,
\[ \lambda = \frac{4n^2}{R_H(n^2 - 4)} \]

5. **Comparing with Given Equation**:
Now, we can compare this with the given equation:
\[ \lambda = c_2 \left( \frac{n^2}{n^2 - 4} \right) \]
By comparing both expressions for \( \lambda \):
\[ c_2 \left( \frac{n^2}{n^2 - 4} \right) = \frac{4n^2}{R_H(n^2 - 4)} \]

6. **Solving for \( c_2 \)**:
From the comparison, we can isolate \( c_2 \):
\[ c_2 = \frac{4}{R_H} \]

### Final Answer:
Thus, the value of \( c_2 \) is:
\[ c_2 = \frac{4}{R_H} \]