Question

The number density of molecules of a gas depends on their distance r from the origin as, $n(r) = n_0e^{-\alpha r^4}$ Then the total number of molecules is proportional to :

• A. $n_0 \alpha^{1/4}$
• B. $n_0 \alpha^{-3}$
• C. $n_0 \alpha^{-3/4}$
• D. $\sqrt{n_0} \alpha^{1/2}$

Answer 1

The Correct Option is C

To determine the proportionality of the total number of molecules in the gas with respect to \(n_0\) and \(\alpha\), we start by considering the given number density of molecules: 

\(n(r) = n_0 e^{-\alpha r^4}\)

The total number of molecules, \(N\), can be found by integrating the number density over the entire space. In spherical coordinates, the volume element is \(dV = r^2 \sin \theta \, dr \, d\theta \, d\phi\). Thus, the total number of molecules is given by:

\(N = \int_0^\infty \int_0^\pi \int_0^{2\pi} n_0 e^{-\alpha r^4} r^2 \sin \theta \, dr \, d\theta \, d\phi\)

To solve this, first integrate over the angular coordinates, which contribute a factor of \(4\pi\) due to the symmetry:

\(\int_0^\pi \sin \theta \, d\theta \int_0^{2\pi} d\phi = 4\pi\)

Thus, the integration reduces to:

\(N = 4\pi n_0 \int_0^\infty r^2 e^{-\alpha r^4} \, dr\)

To evaluate the integral, perform a substitution. Let \(u = \alpha r^4\). Then, \(du = 4\alpha r^3\, dr\) or \(dr = \frac{du}{4\alpha r^3}\).

Substituting \(r = u^{1/4} / \alpha^{1/4}\), we get:

\(N = 4\pi n_0 \int_0^\infty \left(\frac{u^{1/4}}{\alpha^{1/4}}\right)^2 e^{-u} \frac{du}{4 \alpha \left(\frac{u^{1/4}}{\alpha^{1/4}}\right)^3}\)

Simplify the expression:

\(N = \frac{\pi n_0}{\alpha^{3/4}} \int_0^\infty u^{1/4 - 3/4} e^{-u} \, du\)

Recognize this as the Gamma function \(\Gamma(x)\) for \(x = \frac{3}{4}\):

\(\int_0^\infty u^{x-1} e^{-u} \, du = \Gamma(x)\)

The Gamma function is a constant, so the total number of molecules \(N\) is given by:

\(N \propto n_0 \alpha^{-3/4}\)

Thus, the total number of molecules is proportional to \(n_0 \alpha^{-3/4}\), confirming the correct answer:

• \(n_0 \alpha^{-3/4}\)