In the infrared region of the atomic spectrum of hydrogen, a line is obtained at `3802 cm^(-1)`. Calculate the energy of this photon `(h = 6.626 xx 10^(-34) J` sec)`

Explanation

To calculate the energy of the photon corresponding to the line obtained at `3802 cm^(-1)` in the infrared region of the atomic spectrum of hydrogen, we will follow these steps:

### Step 1: Understand the relationship between wave number and wavelength
The wave number (denoted as \( \bar{\nu} \)) is defined as the reciprocal of the wavelength (\( \lambda \)):
\[ \bar{\nu} = \frac{1}{\lambda} \]
Where \( \bar{\nu} \) is in cm\(^{-1}\) and \( \lambda \) is in cm.

### Step 2: Convert wave number to wavelength
Given the wave number \( \bar{\nu} = 3802 \, \text{cm}^{-1} \), we can find the wavelength:
\[ \lambda = \frac{1}{\bar{\nu}} = \frac{1}{3802 \, \text{cm}^{-1}} \approx 0.000263 \, \text{cm} \]
To convert this to meters (since \( 1 \, \text{cm} = 0.01 \, \text{m} \)):
\[ \lambda \approx 0.000263 \, \text{cm} \times 0.01 \, \text{m/cm} = 2.63 \times 10^{-6} \, \text{m} \]

### Step 3: Calculate the frequency
We can use the speed of light (\( c \)) to find the frequency (\( \nu \)):
\[ c = \lambda \cdot \nu \implies \nu = \frac{c}{\lambda} \]
Where \( c = 3 \times 10^8 \, \text{m/s} \):
\[ \nu = \frac{3 \times 10^8 \, \text{m/s}}{2.63 \times 10^{-6} \, \text{m}} \approx 1.14 \times 10^{14} \, \text{Hz} \]

### Step 4: Calculate the energy of the photon
Using Planck's equation:
\[ E = h \cdot \nu \]
Where \( h = 6.626 \times 10^{-34} \, \text{J s} \):
\[ E = 6.626 \times 10^{-34} \, \text{J s} \cdot 1.14 \times 10^{14} \, \text{Hz} \approx 7.56 \times 10^{-20} \, \text{J} \]

### Final Answer
The energy of the photon is approximately:
\[ E \approx 7.56 \times 10^{-20} \, \text{J} \]