Question

What is the average kinetic energy of helium atoms in a region of the solar corona where the temperature is 5.90 x 10^5 K?

Answer 1

Answer: The average kinetic energy of helium atoms is $1.222\times 10^{-17}J$

Explanation:

To calculate the average kinetic energy of the atom, we use the equation:

$K=\frac{3}{2}kT$

where,

K = average kinetic energy = ?

k = Boltzmann constant = $1.3807\times 10^{-23}J/K$

T = temperature = $5.9\times 10^5K$

Putting values in above equation, we get:

$K=\frac{3}{2}\times 1.3807\times 10^{-23}J/K\times 5.9\times 10^5K\\\\K=1.222\times 10^{-17}J$

Hence, the average kinetic energy of helium atoms is $1.222\times 10^{-17}J$