Question

What is the average kinetic energy in joules of hydrogen atoms on the 5500°C surface of the Sun?

a) 2.04 × 10⁻¹⁹ J
b) 2.85 × 10⁻¹⁹ J
c) 3.49 × 10⁻¹⁹ J
d) 4.21 × 10⁻¹⁹ J

Answer 1

To find the average kinetic energy of hydrogen atoms on the surface of the Sun, you convert the temperature to Kelvin and use the equipartition theorem formula involving Boltzmann's constant.

Step-by-step explanation:

To calculate the average kinetic energy of hydrogen atoms on the 5500°C surface of the Sun, we can use the equipartition theorem from thermodynamics. This theorem relates the average kinetic energy of a particle to the temperature of the system according to the formula ½ kT, where k is Boltzmann's constant and T is the temperature in Kelvin. However, first, we need to convert 5500°C to Kelvin (K). The conversion formula is K = °C + 273.15, so T = 5500 + 273.15 = 5773.15 K.

Now, we can calculate the average kinetic energy (E) of a hydrogen atom:

E = ½ kT = ½ (1.38 × 10⁻²³ J/K) (5773.15 K) = (1.38 × 10⁻²³ J/K) (5773.15 K) / 2 = 3.99 × 10⁻²¹ J

However, this does not match any of the given options exactly. We may need to recheck our calculations or the given options could be incorrect. It's crucial that numerical calculations should always be reviewed closely to avoid any small errors that may lead to incorrect answers.