If wavelength of the first line of the Paschen series of hydrogen atom is 720 nm, then the wavelength of the second line of this series i____________nm. (Nearest integer)

Answer

To find the wavelength of the second line of the Paschen series for the hydrogen atom, we can follow these steps:

### Step 1: Understand the Paschen Series
The Paschen series corresponds to electronic transitions where the electron falls to the n=3 energy level. The first line of the Paschen series corresponds to the transition from n=4 to n=3, and the second line corresponds to the transition from n=5 to n=3.

### Step 2: Identify the Given Information
We are given that the wavelength of the first line (from n=4 to n=3) is 720 nm.

### Step 3: Use the Rydberg Formula
The Rydberg formula for the wavelength of light emitted during a transition between energy levels in a hydrogen atom is given by:

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

Where:
- \( R \) is the Rydberg constant (approximately \( 1.097 \times 10^7 \, \text{m}^{-1} \))
- \( n_1 \) is the lower energy level (for the Paschen series, \( n_1 = 3 \))
- \( n_2 \) is the higher energy level (for the first line, \( n_2 = 4 \))

### Step 4: Calculate the Rydberg Constant Using the First Line
Using the given wavelength for the first line (720 nm), we convert it to meters:

\[ \lambda_1 = 720 \, \text{nm} = 720 \times 10^{-9} \, \text{m} \]

Now, we can rearrange the Rydberg formula to solve for \( R \):

\[ \frac{1}{720 \times 10^{-9}} = R \left( \frac{1}{3^2} - \frac{1}{4^2} \right) \]

Calculating the right side:

\[ \frac{1}{3^2} = \frac{1}{9}, \quad \frac{1}{4^2} = \frac{1}{16} \]

So,

\[ \frac{1}{9} - \frac{1}{16} = \frac{16 - 9}{144} = \frac{7}{144} \]

Now substituting back:

\[ \frac{1}{720 \times 10^{-9}} = R \left( \frac{7}{144} \right) \]

Solving for \( R \):

\[ R = \frac{1}{720 \times 10^{-9}} \times \frac{144}{7} \]

### Step 5: Calculate the Wavelength of the Second Line
Now, we apply the Rydberg formula again for the second line (from n=5 to n=3):

\[ \frac{1}{\lambda_2} = R \left( \frac{1}{3^2} - \frac{1}{5^2} \right) \]

Calculating \( \frac{1}{5^2} \):

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

So,

\[ \frac{1}{3^2} - \frac{1}{5^2} = \frac{1}{9} - \frac{1}{25} \]

Finding a common denominator (225):

\[ \frac{25 - 9}{225} = \frac{16}{225} \]

Now substituting back into the Rydberg formula:

\[ \frac{1}{\lambda_2} = R \left( \frac{16}{225} \right) \]

Substituting the value of \( R \) we found earlier:

\[ \frac{1}{\lambda_2} = \frac{1}{720 \times 10^{-9}} \times \frac{144}{7} \times \frac{16}{225} \]

### Step 6: Solve for \( \lambda_2 \)
Now we can calculate \( \lambda_2 \):

\[ \lambda_2 = \frac{1}{\left( \frac{1}{720 \times 10^{-9}} \times \frac{144 \times 16}{7 \times 225} \right)} \]

Calculating this gives us:

\[ \lambda_2 \approx 492.18 \, \text{nm} \]

### Final Answer
Thus, the wavelength of the second line of the Paschen series is approximately **492 nm** (nearest integer).

---