Maxwell-Boltzmann Energy Distribution

The Maxwell-Boltzmann speed distribution tells us how molecular speeds are spread in a gas at equilibrium; often what we actually want is how the kinetic energies are spread — for reaction rates, effusion, evaporation, and any process governed by an energy threshold. The energy distribution $g(E)$ is obtained from the speed distribution by the change of variables $E=\tfrac12 mv^2$ , and the result is strikingly simple and mass-independent:

g(E)=\frac{2}{\sqrt\pi}\,(k_BT)^{-3/2}\,\sqrt{E}\,e^{-E/k_BT}.

Two features stand out. First, all trace of the molecular mass has vanished — a light molecule and a heavy one at the same temperature have identical energy distributions, even though their speed distributions differ. Second, the $\sqrt E$ factor from the density of states competes with the falling exponential to place the peak at a surprisingly low energy. That competition produces the article's central and most misremembered fact: the most probable energy is $E_{\rm mp}=\tfrac12 k_BT$ , while the mean energy is $\langle E\rangle=\tfrac32 k_BT$ . The distribution is skewed — it rises steeply from zero as $\sqrt E$ , peaks at $\tfrac12 k_BT$ , then decays exponentially with a long high-energy tail that pulls the mean up to three times the peak. Confusing "the peak of the energy distribution" with "the average energy" is one of the most common errors in kinetic theory. The energy distribution is the workhorse behind the Arrhenius law: the fraction of molecules with energy above an activation threshold $E_a\gg k_BT$ is set by the exponential tail, $\propto e^{-E_a/k_BT}$ . It also underlies the surprising fact that molecules effusing through a small hole carry more than the average energy of the gas, and it connects directly to the density of states $g(E)\propto\sqrt E$ that reappears throughout quantum statistical mechanics.

Key takeaways

The energy distribution follows from the speed distribution by $E=\tfrac12 mv^2$ : $g(E)=\tfrac{2}{\sqrt\pi}(k_BT)^{-3/2}\sqrt E\,e^{-E/k_BT}$ .
It is mass-independent — molecules of any mass share the same energy distribution at a given $T$ (unlike the speed distribution).
The $\sqrt E$ prefactor is the three-dimensional density of states; it competes with the falling exponential to set the shape.
Most probable energy $E_{\rm mp}=\tfrac12 k_BT$ (from $\mathrm dg/\mathrm dE=0$ ); mean energy $\langle E\rangle=\tfrac32 k_BT$ ; they differ by a factor of three — the distribution is skewed.
Moments come from gamma functions: $\langle E\rangle=\tfrac32 k_BT$ via $\Gamma(5/2)$ , $\langle E^2\rangle=\tfrac{15}{4}(k_BT)^2$ via $\Gamma(7/2)$ , variance $\tfrac32(k_BT)^2$ .
Key idea: because the density of states grows as $\sqrt E$ but the Boltzmann factor falls exponentially, the energy distribution peaks at $\tfrac12 k_BT$ yet averages $\tfrac32 k_BT$ — the long high-energy tail powers all thermally activated processes.

The core insight

Change variables from speed to energy with $E=\tfrac12 mv^2$ , $\mathrm dE=mv\,\mathrm dv$ . The mass cancels completely and the $\sqrt E$ density-of-states factor emerges: $g(E)=\tfrac{2}{\sqrt\pi}(k_BT)^{-3/2}\sqrt E\,e^{-E/k_BT}$ . The peak sits at $\tfrac12 k_BT$ ; the mean sits at $\tfrac32 k_BT$ ; the gap is the skew.

Where this lives today

The energy distribution's high-energy tail is the physics behind essentially every thermally activated rate: chemical reaction rates (the Arrhenius factor $e^{-E_a/k_BT}$ ), thermionic emission of electrons from hot metals, the evaporation of the fastest molecules from a liquid surface (evaporative cooling, the technique that reaches Bose-Einstein condensation), and the fusion rate in stellar cores, where the tail overlaps the quantum-tunneling probability to produce the Gamow peak. In plasma physics and semiconductor device modeling the same $\sqrt E\,e^{-E/k_BT}$ shape determines carrier populations above a band edge. The mass independence of $g(E)$ is why "temperature" is a universal energy scale across chemically different species in the same reactor or the same star.

The physical picture

Two effects shape the energy distribution and pull in opposite directions. The first is the density of states: in three dimensions, the number of velocity directions with a given kinetic energy grows as $\sqrt E$ (the surface area of a sphere of radius $v\propto\sqrt E$ in velocity space). More energy means more ways to have it, so the low-energy region is suppressed by simple counting. The second is the Boltzmann factor $e^{-E/k_BT}$ , which penalizes high energy exponentially. Their product $\sqrt E\,e^{-E/k_BT}$ rises, turns over, and falls. Where does it turn over? Setting $\mathrm d(\sqrt E\,e^{-E/k_BT})/\mathrm dE=0$ gives $E_{\rm mp}=\tfrac12 k_BT$ — a low peak, because the exponential wins early. But the exponential tail is heavy: there are always some molecules far out at high energy, and they drag the average up to $\tfrac32 k_BT$ , three times the peak. This is the signature asymmetry of the energy distribution, and it is the single fact most worth remembering.

Figure

The energy distribution g(E) ∝ √E e^(−E/kT). Punchline: the peak (most probable energy) sits at E = ½kT, but the mean sits at ⟨E⟩ = 3/2 kT — a factor of three apart, because the long high-energy tail pulls the average up.

The payoff figure is the mass independence. Plot the energy distribution for helium and for xenon — a 33-fold mass difference — at the same temperature, and the curves lie exactly on top of one another, even though their speed distributions are wildly different (helium is far faster). Energy, not speed, is the natural currency of thermal equilibrium: equipartition assigns $\tfrac12 k_BT$ per quadratic degree of freedom regardless of mass, and the energy distribution inherits that democracy.

Figure

Payoff: the energy distributions of helium (light) and xenon (heavy) coincide exactly at the same T, though their speed distributions differ enormously. Energy — not speed — is mass-independent in thermal equilibrium.

The core equations

The energy distribution is defined so that $g(E)\,\mathrm dE$ is the fraction of molecules with kinetic energy between $E$ and $E+\mathrm dE$ . It is obtained from the normalized Maxwell-Boltzmann speed distribution $f(v)$ by conservation of probability, $g(E)\,\mathrm dE=f(v)\,\mathrm dv$ , using $E=\tfrac12 mv^2$ .

g(E)=\frac{2}{\sqrt\pi}\,\frac{\sqrt E}{(k_BT)^{3/2}}\,e^{-E/k_BT} \qquad (1)

where: $g(E)$ — energy probability density (units of inverse energy); $E$ — molecular kinetic energy; $k_BT$ — thermal energy scale; the prefactor $2/\sqrt\pi$ enforces normalization. Reading: $\sqrt E$ is the density of states (more ways to have more energy); $e^{-E/k_BT}$ is the Boltzmann suppression; their product is peaked and skewed. Note the mass $m$ is absent. The most probable energy is where the density peaks. It is not the mean, and locating it correctly is the crux of the topic.

\frac{\mathrm d g}{\mathrm dE}=0 \;\Rightarrow\; E_{\rm mp}=\tfrac12 k_BT,\qquad \langle E\rangle=\tfrac32 k_BT \qquad (2)

where: $E_{\rm mp}$ — most probable (modal) energy, from $\mathrm d(\sqrt E\,e^{-E/k_BT})/\mathrm dE=0$ ; $\langle E\rangle$ — mean energy, the equipartition value for three translational degrees of freedom. Their ratio is exactly $3$ : the distribution's skew. The moments all reduce to gamma-function integrals, because $\int_0^\infty E^{s}\,e^{-E/k_BT}\mathrm dE=(k_BT)^{s+1}\Gamma(s+1)$ .

\langle E^n\rangle=(k_BT)^n\,\frac{\Gamma(n+\tfrac32)}{\Gamma(\tfrac32)},\qquad \langle E\rangle=\tfrac32 k_BT,\quad \langle E^2\rangle=\tfrac{15}{4}(k_BT)^2,\quad \mathrm{Var}(E)=\tfrac32(k_BT)^2 \qquad (3)

where: $\Gamma(\tfrac32)=\tfrac{\sqrt\pi}{2}$ , $\Gamma(\tfrac52)=\tfrac{3\sqrt\pi}{4}$ , $\Gamma(\tfrac72)=\tfrac{15\sqrt\pi}{8}$ ; the variance $\tfrac32(k_BT)^2$ gives a relative width $\sqrt{\mathrm{Var}}/\langle E\rangle=\sqrt{2/3}\approx0.82$ , a broad distribution.

Derivation AThe energy distribution from the speed distribution, and the disappearance of the mass

The goal is to transform the Maxwell-Boltzmann speed distribution into the energy distribution by a change of variables, prove that the molecular mass cancels completely, and confirm the normalization $\int_0^\infty g(E)\mathrm dE=1$ using a gamma function. This is the load-bearing step: it explains both why the $\sqrt E$ density-of-states factor appears and why energy — unlike speed — is a mass-blind currency in thermal equilibrium.

Step 1. Start from the normalized speed distribution $f(v)=4\pi\big(\tfrac{m}{2\pi k_BT}\big)^{3/2}v^2 e^{-mv^2/2k_BT}$ . Probability is conserved under the change of variable, so $g(E)=f(v)\,\dfrac{\mathrm dv}{\mathrm dE}$ with $E=\tfrac12 mv^2$ .

Step 2. Compute the Jacobian: $E=\tfrac12 mv^2\Rightarrow \mathrm dE=mv\,\mathrm dv\Rightarrow \dfrac{\mathrm dv}{\mathrm dE}=\dfrac{1}{mv}=\dfrac{1}{\sqrt{2mE}}$ , using $v=\sqrt{2E/m}$ .

Step 3. Substitute $v^2=2E/m$ and the Jacobian into $g(E)=f(v)/(mv)$ : $g(E)=4\pi\big(\tfrac{m}{2\pi k_BT}\big)^{3/2}\cdot\tfrac{2E}{m}\cdot e^{-E/k_BT}\cdot\tfrac{1}{\sqrt{2mE}}$ . Why the mass cancels: collect the mass powers — $m^{3/2}$ from the prefactor, $m^{-1}$ from $2E/m$ , $m^{-1/2}$ from $1/\sqrt{2mE}$ — giving $m^{3/2-1-1/2}=m^0$ . Every $m$ vanishes.

Step 4. Collect the constants: $4\pi\cdot(2\pi)^{-3/2}\cdot 2\cdot 2^{-1/2}=\dfrac{2}{\sqrt\pi}$ , leaving $g(E)=\dfrac{2}{\sqrt\pi}(k_BT)^{-3/2}\sqrt E\,e^{-E/k_BT}$ , equation (1).

Step 5. Verify normalization: $\int_0^\infty g(E)\mathrm dE=\dfrac{2}{\sqrt\pi}(k_BT)^{-3/2}\int_0^\infty E^{1/2}e^{-E/k_BT}\mathrm dE=\dfrac{2}{\sqrt\pi}(k_BT)^{-3/2}(k_BT)^{3/2}\Gamma(\tfrac32)=\dfrac{2}{\sqrt\pi}\cdot\dfrac{\sqrt\pi}{2}=1$ . ✓

Go deeper · Derivation B · 1 minThe peak at ½kT versus the mean at 3/2 kT — the skew, quantified

Derivation BThe peak at ½kT versus the mean at 3/2 kT — the skew, quantified

The goal is to locate the most probable energy by extremizing $g(E)$ , obtain the mean and variance from gamma-function moments, and show explicitly that the peak and mean differ by a factor of three — the defining asymmetry of the energy distribution and the most common conceptual trap in kinetic theory. Getting these two numbers straight is the single most important outcome of the topic.

Step 1. Locate the peak. Extremize the logarithm: $\ln g=\text{const}+\tfrac12\ln E-E/k_BT$ , so $\dfrac{\mathrm d\ln g}{\mathrm dE}=\dfrac{1}{2E}-\dfrac{1}{k_BT}=0$ , giving $E_{\rm mp}=\tfrac12 k_BT$ . Why so low: the $\sqrt E$ rise is gentle ( $\tfrac12\ln E$ ) while the exponential fall is linear in $E$ , so the balance point sits below $k_BT$ .

Step 2. Compute the mean from equation (3): $\langle E\rangle=k_BT\,\dfrac{\Gamma(5/2)}{\Gamma(3/2)}=k_BT\cdot\dfrac{(3/4)\sqrt\pi}{(1/2)\sqrt\pi}=k_BT\cdot\dfrac32=\tfrac32 k_BT$ . This is precisely the equipartition energy of three translational degrees of freedom, as it must be.

Step 3. The ratio is exact: $\langle E\rangle/E_{\rm mp}=(\tfrac32 k_BT)/(\tfrac12 k_BT)=3$ . Why the mean exceeds the peak: the distribution is right-skewed — the exponential tail extends to arbitrarily high energy while the low-energy side is bounded by $E=0$ , so probability mass in the tail drags $\langle E\rangle$ above the mode.

Step 4. Second moment and variance: $\langle E^2\rangle=(k_BT)^2\dfrac{\Gamma(7/2)}{\Gamma(3/2)}=(k_BT)^2\cdot\dfrac{(15/8)\sqrt\pi}{(1/2)\sqrt\pi}=\tfrac{15}{4}(k_BT)^2$ , so $\mathrm{Var}=\langle E^2\rangle-\langle E\rangle^2=\tfrac{15}{4}-\tfrac94=\tfrac{6}{4}=\tfrac32(k_BT)^2$ . The relative width is $\sqrt{3/2}/(3/2)=\sqrt{2/3}\approx0.82$ — a genuinely broad distribution.

Limiting cases

The energy distribution controls behavior in several important regimes.

Case	Result	Meaning
Peak of $g(E)$	$E_{\rm mp}=\tfrac12 k_BT$	Most probable energy; well below the mean.
Mean of $g(E)$	$\langle E\rangle=\tfrac32 k_BT$	Equipartition energy (three translational dof).
Low energy, $E\ll k_BT$	$g(E)\approx\tfrac{2}{\sqrt\pi}(k_BT)^{-3/2}\sqrt E$	Density-of-states dominated; rises as $\sqrt E$ .
High-energy tail, $E\gg k_BT$	$g(E)\propto\sqrt E\,e^{-E/k_BT}$	Arrhenius activation; fraction above $E_a$ is $\propto e^{-E_a/k_BT}$ .
Effusive beam (flux-weighted)	$\langle E\rangle_{\rm beam}=2k_BT$	Extra velocity weighting shifts the mean higher (see effusion).

Common trapCommon pitfall

The number-one error is to confuse the most probable energy ( $\tfrac12 k_BT$ ) with the average energy ( $\tfrac32 k_BT$ ) — they differ by a factor of three, because the energy distribution is right-skewed. A second, related trap: the most probable energy does NOT correspond to the most probable speed. The peak speed satisfies $v_{\rm mp}=\sqrt{2k_BT/m}$ , so $\tfrac12 mv_{\rm mp}^2=k_BT$ , not $\tfrac12 k_BT$ — because the Jacobian $\mathrm dv/\mathrm dE$ reshapes the distribution and shifts the peak. Never equate "peak of $f(v)$ mapped through $E=\tfrac12 mv^2$ " with "peak of $g(E)$ ." Finally, remember $g(E)$ is mass-independent while $f(v)$ is not; do not carry a molecular mass into an energy-distribution calculation.

Worked examples

Example 1 — Fraction of molecules above an activation threshold (enumerate the tail, apply the integral, get a number). A reaction needs each molecule to bring at least $E_a=0.5\,\mathrm{eV}$ of kinetic energy. At $T=500\,\mathrm K$ , what fraction of molecules qualify?

Step 1. Enumerate: the fraction is $\displaystyle F(>E_a)=\int_{E_a}^{\infty}g(E)\,\mathrm dE$ with $g$ from equation (1). Let $x=E/k_BT$ ; then $F=\dfrac{2}{\sqrt\pi}\int_{x_a}^{\infty}\sqrt x\,e^{-x}\,\mathrm dx$ , the upper incomplete gamma $\Gamma(\tfrac32,x_a)/\Gamma(\tfrac32)$ .

Step 2. Apply the threshold: $k_BT=(1.381\times10^{-23})(500)=6.90\times10^{-21}\,\mathrm J=0.0431\,\mathrm{eV}$ , so $x_a=E_a/k_BT=0.5/0.0431=11.6$ .

Step 3. Boundary number: for $x_a\gg1$ the incomplete gamma is $\Gamma(\tfrac32,x_a)\approx\sqrt{x_a}\,e^{-x_a}$ , so $F\approx\dfrac{2}{\sqrt\pi}\sqrt{x_a}\,e^{-x_a}=\dfrac{2}{\sqrt\pi}\sqrt{11.6}\,e^{-11.6}=(1.128)(3.41)(9.1\times10^{-6})=\mathbf{3.5\times10^{-5}}$ . Only about one molecule in thirty thousand clears the barrier — and the exponential $e^{-E_a/k_BT}$ is what makes reaction rates so temperature-sensitive.

Example 2 — Verifying the peak-versus-mean split for nitrogen at room temperature. Take $\mathrm N_2$ at $T=300\,\mathrm K$ ; $k_BT=4.14\times10^{-21}\,\mathrm J=0.0259\,\mathrm{eV}$ .

Step 1. Most probable energy: $E_{\rm mp}=\tfrac12 k_BT=2.07\times10^{-21}\,\mathrm J=\mathbf{0.0129\,eV}$ — the modal molecule carries half the thermal energy.

Step 2. Mean energy: $\langle E\rangle=\tfrac32 k_BT=6.21\times10^{-21}\,\mathrm J=\mathbf{0.0388\,eV}$ — the average is three times the mode.

Step 3. Boundary check: the ratio $\langle E\rangle/E_{\rm mp}=0.0388/0.0129=3.00$ , and the RMS energy spread is $\sqrt{\mathrm{Var}}=\sqrt{3/2}\,k_BT=1.22 k_BT=0.0317\,\mathrm{eV}$ — comparable to the mean itself, confirming a broad, skewed distribution. The peak, the mean, and the width are all order $k_BT$ but numerically distinct — quoting the right one is the whole game.

Origins · 1 minkinetic theory of gases and how this was discovered

kinetic theory of gases

statistical-mechanics concept · Maxwell 1860; Boltzmann 1868–77

The kinetic theory of gases explains macroscopic pressure, temperature, and transport as statistical averages over the chaotic motion of molecules — and its crowning result is the distribution of molecular energies. In the classical gas the energy distribution takes the form $g(E)\propto\sqrt E\,e^{-E/k_BT}$ , the $\sqrt E$ arising from the surface area of a sphere in three-dimensional velocity space. James Clerk Maxwell first derived the velocity distribution (1860), the founding calculation of kinetic theory, and Ludwig Boltzmann generalized it to arbitrary systems in equilibrium (1868–77), giving the Boltzmann factor $e^{-E/k_BT}$ its universal role. The same $\sqrt E$ energy weighting reappears throughout quantum statistical mechanics — in the free-electron gas, the photon gas, and the ideal Bose and Fermi gases — where it multiplies the appropriate occupation function to build every thermodynamic integral.

History

1860

— Maxwell derives the velocity distribution of gas molecules.

1868

— Boltzmann generalizes the distribution to systems in thermal equilibrium.

1877

— Boltzmann connects the distribution to entropy and probability,

S=k_B\ln W

1889

— Arrhenius formulates the activation-energy law

k\propto e^{-E_a/RT}

, powered by the exponential tail.

1920

— Stern's molecular-beam experiments confirm the Maxwell-Boltzmann speed and energy distributions directly.

Going further

References. Pathria & Beale, Statistical Mechanics, 3rd ed., Ch. 6 · Reif, Fundamentals of Statistical and Thermal Physics, Ch. 7 · Kardar, Statistical Physics of Particles, Ch. 4 · Sethna, Statistical Mechanics: Entropy, Order Parameters, and Complexity, Ch. 3 · Landau & Lifshitz, Statistical Physics Part 1, §38. All equations follow from the canonical velocity distribution via the change-of-variables $E=\tfrac12 mv^2$ . Key gamma-function values used throughout: $\Gamma(3/2)=\tfrac12\sqrt\pi$ , $\Gamma(5/2)=\tfrac34\sqrt\pi$ , $\Gamma(7/2)=\tfrac{15}{8}\sqrt\pi$ , which yield $\langle E\rangle=\tfrac32 k_BT$ , $\langle E^2\rangle=\tfrac{15}{4}(k_BT)^2$ , $\mathrm{Var}=\tfrac32(k_BT)^2$ , $E_{\rm mp}=\tfrac12 k_BT$ , effusive beam mean $2k_BT$ , and normalization $\int_0^\infty g(E)\mathrm dE=1$ . Boltzmann constant: $k_B=1.380649\times10^{-23}\,\mathrm{J/K}=8.617\times10^{-5}\,\mathrm{eV/K}$ .

Builds on: Maxwell-Boltzmann Speed Distribution · Maxwell-Boltzmann Velocity Distribution · Classical Ideal Gas · Equipartition Theorem Leads to: Characteristic Speeds: Mean, RMS, and Most Probable · Generalized Equipartition Theorem · Sackur-Tetrode Equation

Why does the energy distribution peak at ½kT and not at the average 3/2 kT?

Because the distribution is skewed. The peak (mode) is where $g(E)$ is largest, found by setting $\mathrm dg/\mathrm dE=0$ , which gives $E_{\rm mp}=\tfrac12 k_BT$ . The mean is the balance point of the whole distribution, and because there is a long exponential tail extending to high energy — while the low side is cut off at $E=0$ — the mean is pulled above the mode, to $\tfrac32 k_BT$ . Both numbers are correct; they just answer different questions. "Most likely energy of a molecule" is $\tfrac12 k_BT$ ; "average energy per molecule" is $\tfrac32 k_BT$ .

Why is the energy distribution mass-independent when the speed distribution is not?

When you change variables from speed to energy, every factor of mass cancels: the $m^{3/2}$ in the speed-distribution prefactor is exactly undone by the $m^{-1}$ from $v^2=2E/m$ and the $m^{-1/2}$ from the Jacobian $\mathrm dv/\mathrm dE=1/\sqrt{2mE}$ . Physically, equilibrium shares energy — not speed — democratically: equipartition gives every molecule $\tfrac32 k_BT$ of translational energy regardless of mass. A heavy molecule reaches that energy by moving slowly and a light one by moving fast, so their speeds differ but their energies are identically distributed.

How does this connect to the Arrhenius law and reaction rates?

The Arrhenius factor $e^{-E_a/RT}$ is the tail of the energy distribution. The fraction of molecules with energy above an activation barrier $E_a$ is $\int_{E_a}^\infty g(E)\mathrm dE$ , and for $E_a\gg k_BT$ this is dominated by the exponential, $\propto e^{-E_a/k_BT}$ (times a slowly varying $\sqrt{E_a}$ ). Because the exponent is large and negative, small changes in temperature change the reacting fraction enormously — which is why a $10\,\mathrm K$ rise can double a reaction rate. The pre-exponential factor and the collision geometry modify the prefactor, but the temperature sensitivity comes straight from the Maxwell-Boltzmann tail.

What is the relationship between g(E) and the density of states in quantum statistics?

They are the same object in classical form. The $\sqrt E$ in $g(E)$ is the density of single-particle states per unit energy for a free particle in three dimensions, $g(\varepsilon)\propto\sqrt\varepsilon$ , which comes from counting the volume of a sphere of radius $|\mathbf p|\propto\sqrt\varepsilon$ in momentum space. In quantum statistical mechanics you multiply exactly this $\sqrt\varepsilon$ density of states by the appropriate occupation function — Bose-Einstein or Fermi-Dirac instead of Boltzmann — to build the thermodynamic integrals for photon gases, electron gases, and Bose gases. The classical energy distribution is the high-temperature limit where the occupation function reduces to $e^{-\varepsilon/k_BT}$ .

Why do molecules effusing through a hole carry more than the average energy?

Effusion samples molecules in proportion to how fast they cross the hole, so faster (higher-energy) molecules are over-represented in the beam. The extra velocity weighting multiplies the bulk distribution by another factor of $v$ , shifting the beam's energy distribution to higher energy: the mean energy of the effusing beam is $2k_BT$ , not the bulk $\tfrac32 k_BT$ . This is a real and measurable bias — it is why molecular beams are "hot" relative to their source and why effusion slightly cools the residual gas (it preferentially removes energetic molecules).

Is the energy distribution the same as the Boltzmann distribution?

Not quite — it is the Boltzmann factor multiplied by the density of states. The bare Boltzmann distribution $P\propto e^{-E/k_BT}$ gives the relative probability of a single microstate of energy $E$ , and it is monotonically decreasing. The energy distribution $g(E)$ gives the probability of finding a molecule with energy in $[E,E+\mathrm dE]$ , which requires counting how many microstates have that energy — the density of states $\sqrt E$ . Multiplying $\sqrt E$ (rising) by $e^{-E/k_BT}$ (falling) produces the peaked, skewed $g(E)$ . Forgetting the density-of-states factor is why some students expect the distribution to peak at $E=0$ .