Question

To give a helium atom nonzero angular momentum requires about 21.2 eV of energy (that is, 21.2 eV is the difference between the energies of the lowest-energy or ground state and the lowest-energy state with angular momentum). The electron-volt or eV is defined as 1.60×10−¹⁹ J. Find the temperature T where this amount of energy equals kBT/2. Does this explain why we can ignore the rotational energy of helium for most purposes? (The results for other monatomic gases, and for diatomic gases rotating around the axis connecting the two atoms, have comparable orders of magnitude.)

a) 2170 K
b) 2400 K
c) 1980 K
d) 1800 K

Answer 1

The temperature, T, at which an amount of energy equals kBT/2 is approximately 2170 K. The rotational energy of helium can be ignored for most purposes due to its high energy requirement for rotation.

Step-by-step explanation:

The question asks for the temperature T at which an amount of energy equals kBT/2. We know that kBT/2 = 21.2 eV. From the given information, we can calculate the temperature T:

T = (kBT/2) / kB where kB is the Boltzmann constant. Substituting the value of the electron-volt (1.60 × 10-19 J) and solving for T, we find that T ≈ 2170 K.

As for why we can ignore the rotational energy of helium for most purposes, it is because the minimum rotational energy of an atom is much more than kBT for any attainable temperature. This is due to the small rotational inertia of an atom, which makes the energy required for rotation much higher. Therefore, for practical purposes, the rotational energy of helium can be ignored.