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Q12.7 (ii):
Estimate the average thermal energy of a helium atom at (ii) the temperature on the surface of the Sun (6000 K),

Solution :

Step 1: Identifying Physical Constants and Parameters

To determine the average thermal energy of a monatomic gas atom (such as Helium), we utilize the Equipartition Theorem of energy.

  • Temperature ($T$): $6000 \text{ K}$
  • Degrees of freedom ($f$) for a monatomic gas: $3$ (translational motion in $x, y, z$ directions) [Per the Law of Equipartition of Energy]
  • Boltzmann Constant ($k_B$): $1.38 \times 10^{-23} \text{ J/K}$

Step 2: Theoretical Formulation

The average thermal energy ($\bar{E}$) of a gas molecule is defined by the contribution of $k_B T / 2$ per degree of freedom. For a monatomic atom with $f = 3$:

$\bar{E} = \frac{f}{2} k_B T$

Substituting $f = 3$:

$\bar{E} = \frac{3}{2} k_B T$ [Derived from the Kinetic Theory of Gases]

Step 3: Numerical Calculation

Substitute the given variables into the derived equation:

$\bar{E} = 1.5 \times (1.38 \times 10^{-23} \text{ J/K}) \times (6000 \text{ K})$

Perform the arithmetic multiplication:

$\bar{E} = 1.5 \times 1.38 \times 6000 \times 10^{-23} \text{ J}$

$\bar{E} = 2.07 \times 6000 \times 10^{-23} \text{ J}$

$\bar{E} = 12420 \times 10^{-23} \text{ J}$

Standardizing the scientific notation:

$\bar{E} = 1.242 \times 10^4 \times 10^{-23} \text{ J}$

$\bar{E} = 1.242 \times 10^{-19} \text{ J}$

Visual Representation of Kinetic Distribution

Average Kinetic Energy ($\bar{E}$) Energy $\bar{E} = 1.24 \times 10^{-19} \text{ J}$

Final Solution: The average thermal energy of a helium atom at 6000 K is $1.24 \times 10^{-19} \text{ J}$.


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