Helium atom is two times heavier than a hydrogen molecule. At 298 K, the average translational kinetic energy of a Helium atom is

Options

A. two times that of hydrogen molecule.
B. same as that of a hydrogen molecule.
C. four times that of a hydrogen molecule.
D. half that of a hydrogen molecule.

Explanation

To find the average translational kinetic energy of a helium atom at 298 K, we can use the formula for the average translational kinetic energy of a gas particle, which is given by:

\[ KE = \frac{3}{2} k T \]

where:
- \(KE\) is the average translational kinetic energy,
- \(k\) is the Boltzmann constant (\(1.38 \times 10^{-23} \, \text{J/K}\)),
- \(T\) is the temperature in Kelvin.

### Step-by-Step Solution:

1. **Identify the temperature**:
We are given the temperature \(T = 298 \, \text{K}\).

2. **Use the formula for kinetic energy**:
The average translational kinetic energy for one atom is given by:
\[ KE = \frac{3}{2} k T \]

3. **Substitute the values**:
We will substitute the values into the equation. First, we need to calculate \(k \times T\):
\[ k = 1.38 \times 10^{-23} \, \text{J/K} \]
\[ T = 298 \, \text{K} \]
Now calculate \(kT\):
\[ kT = 1.38 \times 10^{-23} \, \text{J/K} \times 298 \, \text{K} = 4.11 \times 10^{-21} \, \text{J} \]

4. **Calculate the average translational kinetic energy**:
Now, substitute \(kT\) into the kinetic energy formula:
\[ KE = \frac{3}{2} \times 4.11 \times 10^{-21} \, \text{J} = 6.16 \times 10^{-21} \, \text{J} \]

5. **Final Result**:
The average translational kinetic energy of a helium atom at 298 K is:
\[ KE \approx 6.16 \times 10^{-21} \, \text{J} \]