Consider a gas of particles with mass m in a gravitational field g. The number density n(h) at height h follows the Boltzmann distribution. What is the ratio of the number of particles in the height range 0 to H to the number of particles in the range H to infinity?

A. ( mgHkT) - 1 — The number of particles is the integral of the density function over the height. Integrating n(h) from 0 to Hand H to infinity gives the ratio (1 - exp(-mgH/kT)) / exp(-mgH/kT), which simplifies to exp(mgH/kT) - 1.

JEE Main · PhysicsHard

$Consider a gas of particles with mass $m$ in a gravitational field $g$. The number density $n(h) = n_0 \exp(-mgh/kT)$. What is the ratio of the number of particles in the height range $0$ to $H$ to the number of particles in the range $H$ to $\infty$?$

A. $\exp(\frac{mgH}{kT}) - 1$
B. $1 - \exp(-\frac{mgH}{kT})$
C. $\exp(-\frac{mgH}{kT})$
D. $\exp(\frac{mgH}{kT})$

Show correct answer & step-by-step solution

Correct answer: A — $\exp(\frac{mgH}{kT}) - 1$

Solution

Number

N = \int n (h) d h

. Ratio is

\frac{\int _{0}^{H} e ^{- m g h / k T} d h}{\int _{H}^{\infty} e ^{- m g h / k T} d h} = \frac{1 - e ^{- m g H / k T}}{e ^{- m g H / k T}} = e^{m g H / k T} - 1

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