Calculate the wave number and energy of the spectral line emitted by an excited hydrogen atom corresponding to `n_(1)= 3 and n_2 = 5.`

Answer

To solve the problem of calculating the wave number and energy of the spectral line emitted by an excited hydrogen atom corresponding to \( n_1 = 3 \) and \( n_2 = 5 \), we will follow these steps:

### Step 1: Calculate the Wave Number

The formula for the wave number (\( \bar{\nu} \)) is given by the Rydberg formula:

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

Where:
- \( R \) is the Rydberg constant, approximately \( 1.097 \times 10^5 \, \text{cm}^{-1} \)
- \( n_1 = 3 \)
- \( n_2 = 5 \)

Substituting the values into the formula:

\[ \bar{\nu} = 1.097 \times 10^5 \left( \frac{1}{3^2} - \frac{1}{5^2} \right) \]

Calculating \( \frac{1}{3^2} \) and \( \frac{1}{5^2} \):

\[ \frac{1}{3^2} = \frac{1}{9} \approx 0.1111 \]
\[ \frac{1}{5^2} = \frac{1}{25} = 0.04 \]

Now, substituting these values back into the equation:

\[ \bar{\nu} = 1.097 \times 10^5 \left( 0.1111 - 0.04 \right) \]
\[ \bar{\nu} = 1.097 \times 10^5 \times 0.0711 \]
\[ \bar{\nu} \approx 7800 \, \text{cm}^{-1} \]

### Step 2: Calculate the Energy of the Spectral Line

The energy (\( E \)) of the emitted photon can be calculated using the formula:

\[ E = h c \bar{\nu} \]

Where:
- \( h \) is Planck's constant, approximately \( 6.626 \times 10^{-34} \, \text{J s} \)
- \( c \) is the speed of light, approximately \( 3.00 \times 10^8 \, \text{m/s} \)
- \( \bar{\nu} \) is the wave number calculated in Step 1, converted to meters (1 cm = 0.01 m):

\[ \bar{\nu} = 7800 \, \text{cm}^{-1} = 7800 \times 100 = 780000 \, \text{m}^{-1} \]

Now substituting the values into the energy formula:

\[ E = (6.626 \times 10^{-34}) \times (3.00 \times 10^8) \times (780000) \]

Calculating:

\[ E \approx 6.626 \times 10^{-34} \times 3.00 \times 10^8 \times 780000 \]

Calculating step by step:

1. Calculate \( 6.626 \times 3.00 = 19.878 \times 10^{-34} \)
2. Calculate \( 19.878 \times 780000 = 15.5 \times 10^{-34} \)
3. Multiply by \( 10^8 \) to adjust the exponent:

\[ E \approx 1.55 \times 10^{-19} \, \text{J} \]

### Final Results

- The wave number of the spectral line is approximately \( 7.8 \times 10^3 \, \text{cm}^{-1} \).
- The energy of the spectral line is approximately \( 1.55 \times 10^{-19} \, \text{J} \).