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

Options

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

Answer

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

### Step 1: Identify the Series and the Transition
The Balmer series corresponds to transitions where the final energy level (n1) is 2. The third line in the Balmer series corresponds to a transition from n2 = 5 to n1 = 2.

### Step 2: Use the Rydberg Formula
The Rydberg formula for the wavelength (λ) of the emitted light during a transition is given by:

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

Where:
- \( R \) is the Rydberg constant (approximately \( 1.097 \times 10^7 \, \text{m}^{-1} \))
- \( Z \) is the atomic number (for hydrogen, \( Z = 1 \))
- \( n_1 \) is the lower energy level (2 for Balmer series)
- \( n_2 \) is the upper energy level (5 for the third line)

### Step 3: Substitute the Values into the Formula
Substituting the values into the Rydberg formula:

\[ \frac{1}{\lambda} = 1.097 \times 10^7 \cdot 1^2 \left( \frac{1}{2^2} - \frac{1}{5^2} \right) \]

Calculating the fractions:

\[ \frac{1}{2^2} = \frac{1}{4} = 0.25 \]
\[ \frac{1}{5^2} = \frac{1}{25} = 0.04 \]

Now, substituting these values:

\[ \frac{1}{\lambda} = 1.097 \times 10^7 \left( 0.25 - 0.04 \right) \]
\[ \frac{1}{\lambda} = 1.097 \times 10^7 \cdot 0.21 \]

### Step 4: Calculate \( \frac{1}{\lambda} \)
Now, calculate \( 1.097 \times 10^7 \cdot 0.21 \):

\[ \frac{1}{\lambda} = 2.3037 \times 10^6 \, \text{m}^{-1} \]

### Step 5: Find \( \lambda \)
To find \( \lambda \), take the reciprocal:

\[ \lambda = \frac{1}{2.3037 \times 10^6} \approx 4.344 \times 10^{-7} \, \text{m} = 434.4 \, \text{nm} \]

### Conclusion
The wavelength of the third line of the Balmer series for a hydrogen atom is approximately **434.4 nm**.