The wavelength of the third line of the Balmer series for a hydrogen atom is -

Options

A. `(21)/(100R_(infty))`
B. `(100)/(21R_(infty))`
C. `(21R_(infty))/(100)`
D. `(100 R_(infty))/(21)`

Answer

To find the wavelength of the third line of the Balmer series for a hydrogen atom, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Balmer Series**: The Balmer series corresponds to electronic 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. **Identify the Transition for the Third Line**: The third line of the Balmer series corresponds to the transition from n=5 to n=2. Therefore, we have:
- \( n_1 = 2 \) (lower energy level)
- \( n_2 = 5 \) (higher energy level)

3. **Use the Rydberg Formula**: The Rydberg formula for the wavelength of light emitted during these transitions is given by:
\[ \frac{1}{\lambda} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
where \( R_H \) is the Rydberg constant.

4. **Substitute the Values**: Now substitute \( n_1 = 2 \) and \( n_2 = 5 \) into the formula:
\[ \frac{1}{\lambda} = R_H \left( \frac{1}{2^2} - \frac{1}{5^2} \right) \]
This simplifies to:
\[ \frac{1}{\lambda} = R_H \left( \frac{1}{4} - \frac{1}{25} \right) \]

5. **Calculate the Right Side**: To combine the fractions, find a common denominator (which is 100):
\[ \frac{1}{4} = \frac{25}{100}, \quad \frac{1}{25} = \frac{4}{100} \]
Thus,
\[ \frac{1}{\lambda} = R_H \left( \frac{25}{100} - \frac{4}{100} \right) = R_H \left( \frac{21}{100} \right) \]

6. **Solve for Wavelength**: Rearranging gives:
\[ \lambda = \frac{100}{21 R_H} \]

7. **Final Result**: The wavelength of the third line of the Balmer series for a hydrogen atom is:
\[ \lambda = \frac{100}{21 R_H} \]