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Exercises · 12.7

Q.Estimate the average thermal energy of a helium atom at

(i) room temperature (27 ∘C27\ ^\circ\text{C}),
(ii) the temperature on the surface of the Sun (6000 K6000\ \text{K}),
(iii) the temperature of 1010 million kelvin (the typical core temperature in the case of a star).
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The average thermal energy of a monatomic gas like helium is given by 32kBT\frac{3}{2} k_B T, independent of atomic mass. At room temperature it is about 6.21×10−21 J6.21 \times 10^{-21}\ \text{J}, at the Sun’s surface about 1.24×10−19 J1.24 \times 10^{-19}\ \text{J}, and at stellar core temperatures about 2.07×10−16 J2.07 \times 10^{-16}\ \text{J}.

The Kinetic Theory of Gases tells us that temperature is a measure of the average random kinetic energy of the molecules in a gas. For a monatomic gas — one where each atom moves independently without rotation or vibration — the only energy store is translational kinetic energy. The equipartition theorem states that each quadratic degree of freedom contributes 12kBT\frac{1}{2} k_B T to the average energy per particle. A monatomic atom has three translational degrees of freedom (motion along xx, yy, and zz), so its average thermal energy is:

⟨E⟩=32kBT\langle E \rangle = \frac{3}{2} k_B T

where kB=1.38×10−23 J/Kk_B = 1.38 \times 10^{-23}\ \text{J/K} is Boltzmann’s constant, and TT is the absolute temperature in kelvin. This result is independent of the mass of the atom — helium, neon, or argon all have the same average thermal energy at the same temperature. The mass only affects the speed distribution, not the average energy.

Now we apply this formula to each case. Remember to convert Celsius to kelvin: T(K)=T(∘C)+273T(\text{K}) = T(^\circ\text{C}) + 273.

  1. Room temperature (27 ∘C27\ ^\circ\text{C}) Convert: T=27+273=300 KT = 27 + 273 = 300\ \text{K}. Then:

⟨E⟩=32×(1.38×10−23)×300\langle E \rangle = \frac{3}{2} \times (1.38 \times 10^{-23}) \times 300

=32×4.14×10−21=6.21×10−21 J= \frac{3}{2} \times 4.14 \times 10^{-21} = 6.21 \times 10^{-21}\ \text{J}

This is a tiny amount of energy — about 0.039 eV0.039\ \text{eV} (since 1 eV=1.6×10−19 J1\ \text{eV} = 1.6 \times 10^{-19}\ \text{J}). At everyday temperatures, atomic energies are on the order of hundredths of an electronvolt.

  1. Surface of the Sun (6000 K6000\ \text{K}) No conversion needed — already in kelvin.

⟨E⟩=32×(1.38×10−23)×6000\langle E \rangle = \frac{3}{2} \times (1.38 \times 10^{-23}) \times 6000

=32×8.28×10−20=1.242×10−19 J= \frac{3}{2} \times 8.28 \times 10^{-20} = 1.242 \times 10^{-19}\ \text{J}

That’s about 0.78 eV0.78\ \text{eV}. Still modest, but enough to excite some atomic transitions — this is why the Sun’s spectrum shows absorption lines from excited atoms.

  1. Stellar core (107 K10^7\ \text{K})

⟨E⟩=32×(1.38×10−23)×107\langle E \rangle = \frac{3}{2} \times (1.38 \times 10^{-23}) \times 10^7

=32×1.38×10−16=2.07×10−16 J= \frac{3}{2} \times 1.38 \times 10^{-16} = 2.07 \times 10^{-16}\ \text{J} …

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