Calculate the average kinetic energy of hydrogen molecule at `0^@ C`. Given `k_(B) = 1.38 xx 10^(-23) JK^(-1)`.

Explanation

To calculate the average kinetic energy of a hydrogen molecule at \(0^\circ C\), we can follow these steps:

### Step 1: Understand the Formula for Average Kinetic Energy
The average kinetic energy (\(KE\)) of a molecule in a gas can be calculated using the formula:
\[ KE = \frac{F}{2} k_B T \]
where:
- \(F\) is the degrees of freedom of the molecule,
- \(k_B\) is the Boltzmann constant,
- \(T\) is the absolute temperature in Kelvin.

### Step 2: Identify the Degrees of Freedom for Hydrogen Molecule
For a diatomic gas like hydrogen (\(H_2\)), the degrees of freedom (\(F\)) is 5. This includes:
- 3 translational degrees of freedom (motion in x, y, and z directions),
- 2 rotational degrees of freedom (rotation about two axes).

### Step 3: Convert Temperature to Kelvin
The temperature given is \(0^\circ C\). We need to convert this to Kelvin:
\[ T = 0 + 273 = 273 \, K \]

### Step 4: Substitute Values into the Formula
Now, we can substitute the values into the formula:
- \(F = 5\)
- \(k_B = 1.38 \times 10^{-23} \, J/K\)
- \(T = 273 \, K\)

Substituting these values into the kinetic energy formula:
\[ KE = \frac{5}{2} \times (1.38 \times 10^{-23}) \times 273 \]

### Step 5: Calculate the Average Kinetic Energy
Now, we perform the calculation:
\[ KE = \frac{5}{2} \times 1.38 \times 10^{-23} \times 273 \]
Calculating the multiplication:
\[ KE = \frac{5 \times 1.38 \times 273}{2} \times 10^{-23} \]
Calculating \(5 \times 1.38 \times 273\):
\[ 5 \times 1.38 = 6.9 \]
\[ 6.9 \times 273 \approx 1889.7 \]
Now, divide by 2:
\[ \frac{1889.7}{2} \approx 944.85 \]
Thus, we have:
\[ KE \approx 944.85 \times 10^{-23} \, J \]
This can be approximated as:
\[ KE \approx 9.45 \times 10^{-21} \, J \]

### Final Answer
The average kinetic energy of a hydrogen molecule at \(0^\circ C\) is approximately:
\[ KE \approx 9.45 \times 10^{-21} \, J \]