Question

What is the average kinetic energy (in J) of helium atoms in a region of the solar corona where the temperature is 5.70 ✕ 10^5 K?

Answer 1

Given:

The temperature is

`T=\text{ 5.7}*10^5\text{ K}`

Required: Average kinetic energy

Step-by-step explanation:

The average kinetic energy can be calculated by the formula

`E_k=(3)/(2)k_BT`

Here, the Boltzmann constant is

`k_B=\text{ 1.38}*10^(-23)\text{ J/K}`

On substituting the values, the average kinetic energy will be

`\begin{gathered} E_k=(3)/(2)k_BT \\ =(3)/(2)*1.38*10^(-23)*5.7*10^5 \\ =1.1799*10^(-17)J \end{gathered}`

Final Answer: The average kinetic energy of the helium atoms is 1.1799 x 10^(-17) J.