A box contains about 5.64 x 1021 hydrogen
atoms at room temperature 21°C.
Find the thermal energy of these atoms.
Answer in units of J.

Explanation:

Given that,

The temperature of atom is 21°C

θ = 21 + 273 = 294K.

Then, from kinetic theory,

The kinetic energy of individual elements is

K.E = 2kT/3

k is Boltzmann constant

k = 1.38 × 10^-23 J/K•particle

Then,

K.E = 2 × 1.38 × 10^-23 × 294 / 3

K.E = 2.7048 × 10^-21 J/particle

Each particle of hydrogen is 5.64 × 10²¹ particle

Total K.E = 2.7048 × 10^-21 × 5.64 × 10²¹

Total K.E = 15.26 J

c. rb

explanation:

the atomic radius of a chemical element is a measure of the size of its atoms, usually the mean or typical distance from the center of the nucleus to the boundary of the surrounding shells of electrons.

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