The temperature of the surface of the Sun is 5500°C. a. What is the average kinetic energy, in joules, of hydrogen atoms on the surface of the Sun?
b. What is the average kinetic energy, in joules, of helium atoms in a region of the solar corona where the temperature is 6.00 times 10^5 K?

a. the average kinetic energy of hydrogen atoms is  1.20 × 10^-19J

b. the average kinetic energy of helium atoms is 1.24 × 10^-17J

Explanation:

The computation is shown below;

As we know that

Kinetic energy = 3 ÷ 2 kT

where,

K = Boltzmann constant

And, T = Temperature

a. Now the temperature in kelvin is

T = (5,500 × (°C ÷ K) + 273.15 K)

= 5773.15 K

As

Kinetic energy = 3 ÷ 2 kT

So now 1.38 × 10^-23 J/K for K would be substituted and 5773.15 K for Temperature T

Now Kinetic energy is

= 3 ÷ 2 (1.38 × 10^-23 J/K) ( 5773.15 K)

= 1.20 × 10^-19J

hence, the average kinetic energy of hydrogen atoms is  1.20 × 10^-19J

b. As

Kinetic energy = 3 ÷ 2 kT

now 1.38 × 10^-23 J/K for K would be substituted and 6 × 10^5K for Temperature T

Now Kinetic energy is

= 3 ÷ 2 (1.38 × 10^-23 J/K) (6 × 10^5K )

= 1.24 × 10^-17J

hence, the average kinetic energy of helium atoms is 1.24 × 10^-17J

low freq = 100 mhz

low wavelength =101.7m

length = 3x10^8/freq

compass can be used day and night, cloudy or sunny