Estimate the average thermal energy of a helium atom at (i) room temperature `27^(@)C` and (ii) the temperature of the surface of the sun ( 6000K)

Explanation

To estimate the average thermal energy of a helium atom at two different temperatures, we can use the formula for thermal energy:

\[ E = \frac{3}{2} k_B T \]

where:
- \( E \) is the average thermal energy,
- \( k_B \) is the Boltzmann constant, approximately \( 1.38 \times 10^{-23} \, \text{J/K} \),
- \( T \) is the temperature in Kelvin.

### Part (i): Room Temperature (27°C)

1. **Convert the temperature from Celsius to Kelvin**:
\[ T = 27^\circ C + 273 = 300 \, K \]

2. **Substitute the values into the formula**:
\[ E = \frac{3}{2} \times (1.38 \times 10^{-23} \, \text{J/K}) \times (300 \, K) \]

3. **Calculate the thermal energy**:
\[ E = \frac{3}{2} \times 1.38 \times 10^{-23} \times 300 \]
\[ E = 6.21 \times 10^{-21} \, \text{J} \]

### Part (ii): Temperature of the Surface of the Sun (6000 K)

1. **Use the given temperature directly**:
\[ T = 6000 \, K \]

2. **Substitute the values into the formula**:
\[ E = \frac{3}{2} \times (1.38 \times 10^{-23} \, \text{J/K}) \times (6000 \, K) \]

3. **Calculate the thermal energy**:
\[ E = \frac{3}{2} \times 1.38 \times 10^{-23} \times 6000 \]
\[ E = 1.24 \times 10^{-19} \, \text{J} \]

### Final Answers:
- Average thermal energy of a helium atom at room temperature (27°C): \( 6.21 \times 10^{-21} \, \text{J} \)
- Average thermal energy of a helium atom at the surface of the Sun (6000 K): \( 1.24 \times 10^{-19} \, \text{J} \)