The number density of molecules of a gas depends on their distance r from the origin as, $$n(r) = n_0 e^{-\alpha r^4}$$. Then the numer of molecules is proportional to:

We are told that the number density (molecules per unit volume) at a distance $$r$$ from the origin is $$n(r)=n_0 e^{-\alpha r^4}$$. To obtain the total number of molecules $$N$$ present in the whole space, we must integrate this density over the entire volume.

Because the density depends only on the radial distance, the gas is spherically symmetric. In spherical polar coordinates the volume element is $$dV = 4\pi r^2\,dr$$. Hence, using the definition $$N=\displaystyle\int_{\text{all space}} n(r)\,dV$$, we write

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

Now we evaluate the integral $$I=\displaystyle\int_0^{\infty} r^2 e^{-\alpha r^4}\,dr.$$ To remove the exponential’s quartic argument, we perform the substitution $$x=\alpha r^4.$$ Then $$r = \left(\dfrac{x}{\alpha}\right)^{1/4}, \qquad dr = \dfrac{1}{4}\,\alpha^{-1/4}\,x^{-3/4}\,dx.$$

We also need $$r^2$$ in terms of $$x$$:

$$ r^2 = \left(\dfrac{x}{\alpha}\right)^{1/2} = x^{1/2}\,\alpha^{-1/2}. $$

Substituting $$r^2$$ and $$dr$$ into $$I$$ gives

$$ I = \int_0^{\infty} \Bigl(x^{1/2}\alpha^{-1/2}\Bigr)\, e^{-x}\, \Bigl(\frac{1}{4}\alpha^{-1/4}x^{-3/4}\Bigr)\,dx. $$

Collecting powers of $$x$$ and $$\alpha$$:

$$ x^{1/2}x^{-3/4}=x^{-1/4}, \qquad \alpha^{-1/2}\alpha^{-1/4}= \alpha^{-3/4}. $$

Thus

$$ I = \frac{1}{4}\,\alpha^{-3/4}\int_0^{\infty} x^{-1/4}e^{-x}\,dx. $$

The remaining integral is a standard Gamma-function. Recall the formula $$\displaystyle\Gamma(s)=\int_0^{\infty}x^{\,s-1}e^{-x}\,dx.$$ Here $$x^{-1/4}=x^{\,(-1/4)}=x^{\,\,(3/4-1)}$$, so $$s=\frac{3}{4}$$. Therefore

$$ \int_0^{\infty}x^{-1/4}e^{-x}\,dx = \Gamma\!\left(\frac{3}{4}\right), $$

a pure numerical constant, independent of $$\alpha$$ and $$n_0$$. Consequently,

$$ I = \frac{1}{4}\,\Gamma\!\left(\frac{3}{4}\right)\,\alpha^{-3/4}. $$

Returning to the expression for $$N$$, we substitute this result:

$$ N = 4\pi n_0\,I = 4\pi n_0\left[\frac{1}{4}\Gamma\!\left(\frac{3}{4}\right)\alpha^{-3/4}\right] = \pi\,\Gamma\!\left(\frac{3}{4}\right)\,n_0\,\alpha^{-3/4}. $$

The factors $$\pi$$ and $$\Gamma\!\left(\frac{3}{4}\right)$$ are fixed numerical constants, so we conclude that

$$ N \propto n_0\,\alpha^{-3/4}. $$

Hence, the correct answer is Option C.