Answered

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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?

Sagot :

Given:

The temperature is

[tex]T=\text{ 5.7}\times10^5\text{ K}[/tex]

Required: Average kinetic energy

Explanation:

The average kinetic energy can be calculated by the formula

[tex]E_k=\frac{3}{2}k_BT[/tex]

Here, the Boltzmann constant is

[tex]k_B=\text{ 1.38}\times10^{-23}\text{ J/K}[/tex]

On substituting the values, the average kinetic energy will be

[tex]\begin{gathered} E_k=\frac{3}{2}k_BT \\ =\frac{3}{2}\times1.38\times10^{-23}\times5.7\times10^5 \\ =1.1799\times10^{-17}J \end{gathered}[/tex]

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

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