Question bank — Maxwell-Boltzmann distribution — full derivation

True. $g(v_x)\propto e^{-m{v}_x^2/2k_{B}T}$ mein koi $v_x^2$-prefactor nahi hai, toh ye ek pure Gaussian hai jiska maximum $v_x=0$ pe hai — kisi bhi ek component ki most likely value us axis pe zero motion hai.

False. $f(v)$ mein extra factor $v^2$ hota hai, jo $v=0$ pe vanish kar jaata hai; peak bahar khich jaata hai most probable speed $v_p=\sqrt{2k_{B}T/m}$ pe. Zero speed matlab ek shell jiska radius zero hai, jisme koi velocity vectors nahi hote.

False. $v_p=\sqrt{2k_{B}T/m}\propto \sqrt{T}$, toh $T$ double karne se $v_p$ sirf $\sqrt{2}\approx 1.41$ se multiply hoti hai, $2$ se nahi. Speed temperature ke square root ke saath scale karti hai.

False. Heavier matlab chhota $v_p\propto 1/\sqrt{m}$, toh curve left shift hoti hai aur narrower aur taller ho jaati hai (ye low speeds ki taraf squeeze hoti hai, spread out nahi). Lighter gases broad aur fast wali hoti hain.

True. Unka ratio $\sqrt{2}:\sqrt{8/\pi }:\sqrt{3}$ fixed hai aur $T$ aur $m$ se independent hai; long high-speed tail hamesha mean speed aur rms speed ko peak ke right side pe le jaati hai.

False. Area hamesha ${\int }_0^{\infty }f(v)dv=1$ hota hai (ye ek probability hai). Heating curve ko move aur reshape karti hai lekin total area kabhi nahi badlta — ek taller peak forced narrow hoti hai aur vice versa.

False. $⟨v_x⟩=0$ sirf yeh kehta hai ki leftward aur rightward motion symmetry se cancel hote hain. Speed average par kabhi zero nahi hoti: mean speed $⟨v⟩=\sqrt{8k_{B}T/\pi m}>0$ kyunki speed negative nahi ho sakti.

False. $g(v_x)$ $0$ ke baare mein symmetric hai, lekin $f(v)$ sirf $v\ge 0$ pe rehti hai aur iska ek long right tail hai — ye skewed hai, aur yahi exact reason hai ki teen characteristic speeds alag hote hain.

$v^2$ geometric factor missing hai. Boltzmann factor ek velocity state ka weight deta hai; speed $4\pi v^2$ states ko us magnitude ke saath lump karta hai, toh $f(v)\propto v^2{e}^{-m{v}^2/2k_{B}T}$.

Rms speed $\sqrt{⟨v^2⟩}$ hai, yaani squares ke mean ka square root, mean itself nahi. Kyunki squaring large values ko emphasize karta hai, $\sqrt{⟨v^2⟩}>⟨v⟩$ kisi bhi spread-out distribution ke liye.

Units mixed hain. Per-molecule mass $m$ (kg mein) ko $k_B$ ke saath use karo, ya molar mass $M$ (kg/mol mein) ko gas constant $R=N_A{k}_B$ ke saath: $v_p=\sqrt{2RT/M}$. $k_B$ (ek per-molecule constant) ko per-mole mass ke saath kabhi pair mat karo.

Hum unhe integrate nahi karte; hum teen component densities ko multiply karte hain aur phir $d{v}_{x}d{v}_{y}d{v}_z$ ko shell volume $4\pi v^{2}dv$ ke roop mein re-express karte hain. Integration sirf baad mein aati hai, averages nikalne ke liye.

$v_p$ mode (peak location) hai, median nahi. Right tail ki wajah se, median $v_p$ se upar hoti hai, toh zyada se aadhe molecules $v_p$ se faster move karte hain — question ka premise galat hai. (Mean speed $⟨v⟩$ aur bhi right mein hai, median se bhi upar.)

Negative $b$ se $e^{+b{v}_x^2}$ milta hai, jo grow karta rehta hai without bound aur normalize nahi ho sakta (iska integral diverge ho jaata hai). $b>0$ is requirement se forced hai ki $\int gd{v}_x=1$ ho.

Independence force karta hai ki $g(v_x)g(v_y)g(v_z)$ sirf sum $v_x^2+v_y^2+v_z^2$ par depend kare; log lene par, $\ln g(v_x)$ ko $v_x^2$ mein linear hona chahiye. $v_x^4$ term us sum ki function mein add up nahi hoga. Dekho Gaussian integrals.

Components statistically independent aur isotropic hote hain, jisse joint distribution factorize ho sakti hai — ek badi simplification. Speed teeno ko mix karta hai, toh shuru mein koi factorization exploit karne ko nahi milti.

Functional form ek unknown, $b$, chhod deta hai. Equipartition har translational degree of freedom ko $\frac{1}{2}{k}_{B}T$ se pin karta hai, jo $⟨v_x^2⟩=k_{B}T/m$ aur hence $b=m/2k_{B}T$ deta hai — yahi temperature inject karta hai.

$v^2$ shell-volume factor sabse bada killer hai: $v=0$ pe $v$ radius ki sphere ki koi surface nahi hai, toh velocity vectors ke liye koi "jagah" nahi hai, Boltzmann weight se koi farak nahi padta. Dekho Kinetic theory of gases.

Same $T$ ke liye, chhota $m$ bada mean speed $⟨v⟩\propto 1/\sqrt{m}$ deta hai, aur effusion rate mean speed ke saath scale karti hai — yahi Graham's law ki root hai.

Ye assume karta hai ki har velocity state Boltzmann weight $e^{-E/k_{B}T}$ carry karta hai bina kisi occupation limit ke. Jab particles states mein crowd hote hain, quantum statistics le leti hai — dekho Maxwell-Boltzmann vs Fermi-Dirac vs Bose-Einstein aur Boltzmann factor and partition function.

Dono speeds mein same $\sqrt{k_{B}T/m}$ factor hota hai, jo ratio mein cancel ho jaata hai, pure number $\sqrt{3/2}$ chhod ke. MB ki shape universal hai; sirf iska horizontal scale $T$ aur $m$ par depend karta hai.

Curve $v=0$ ki taraf collapse ho jaati hai: $v_p\propto \sqrt{T}\to 0$ aur distribution zero speed pe infinitely narrow spike ban jaati hai — saare molecules frozen, vanishing thermal energy ke consistent.

Peak $v_p\propto \sqrt{T}$ arbitrarily high speed ki taraf chali jaati hai aur curve indefinitely flatten aur broaden hoti jaati hai; har finite speed vanishingly probable ho jaati hai jab probability infinity tak spread hoti hai.

Nahi. Har $v>0$ ke liye $v^2>0$ aur $e^{-m{v}^2/2k_{B}T}>0$ dono hain, toh $f(v)>0$ har jagah $(0,\infty )$ par — kisi bhi positive speed ki, chahe kitni bhi badi ho, hamesha nonzero chance hoti hai.

Exactly zero, jaise kisi bhi continuous distribution ke single point ki. Sirf ranges ki nonzero probability hoti hai: $f(v)dv$ ek interval par ek fraction hai, ek point par nahi.

Shell ban jaati hai ek ring jiska circumference $2\pi v$ hai, toh geometric factor $v^1$ hota hai $v^2$ nahi, jo $f(v)\propto v{e}^{-m{v}^2/2k_{B}T}$ deta hai. Peak aur characteristic speeds shift karte hain kyunki "room" factor ki dimension badal gayi.

Nahi — teen identical Gaussians mein factorization isotropy par depend karti thi. Ek preferred axis ke saath, $g(v_z)$ $g(v_x)$ se alag hoti aur clean $4\pi v^2$ shell (jisne direction-independence assume ki thi) apply nahi hoti.