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 total number of molecules is proportional to :

Options

A. `n_(0) alpha^(-3//4)`
B. `sqrt(n_(0) alpha^(½)`
C. `n_(0) alpha^(a//4)`
D. `n_(0)alpha^(-3)`

Answer

To solve the problem, we need to find the total number of molecules, \( N \), given the number density \( n(r) = n_0 e^{-\alpha r^4} \). The total number of molecules can be calculated by integrating the number density over the entire volume.

### Step-by-Step Solution:

1. **Understanding the Volume Element**:
The volume element in spherical coordinates is given by:
\[ dv = 4\pi r^2 dr \]

2. **Setting Up the Integral**:
The total number of molecules \( N \) can be expressed as:
\[ N = \int_0^\infty n(r) \, dv \]
Substituting for \( n(r) \) and \( dv \):
\[ N = \int_0^\infty n_0 e^{-\alpha r^4} (4\pi r^2) \, dr \]
This simplifies to:
\[ N = 4\pi n_0 \int_0^\infty r^2 e^{-\alpha r^4} \, dr \]

3. **Changing Variables**:
To solve the integral \( \int_0^\infty r^2 e^{-\alpha r^4} \, dr \), we can use a substitution. Let:
\[ u = \alpha r^4 \quad \Rightarrow \quad du = 4\alpha r^3 \, dr \quad \Rightarrow \quad dr = \frac{du}{4\alpha r^3} \]
From the substitution, we have \( r^2 = \left(\frac{u}{\alpha}\right)^{\frac{1}{4}} \) and \( r^3 = \left(\frac{u}{\alpha}\right)^{\frac{3}{4}} \).

4. **Changing Limits**:
When \( r = 0 \), \( u = 0 \) and when \( r \to \infty \), \( u \to \infty \).

5. **Substituting in the Integral**:
The integral becomes:
\[ \int_0^\infty r^2 e^{-\alpha r^4} \, dr = \int_0^\infty \left(\frac{u}{\alpha}\right)^{\frac{1}{4}} e^{-u} \frac{du}{4\alpha \left(\frac{u}{\alpha}\right)^{\frac{3}{4}}} \]
Simplifying this gives:
\[ = \frac{1}{4\alpha} \int_0^\infty \frac{u^{\frac{1}{4}}}{\alpha^{\frac{1}{4}}} e^{-u} \, du = \frac{1}{4\alpha^{\frac{3}{4}}} \int_0^\infty u^{\frac{1}{4}} e^{-u} \, du \]

6. **Using the Gamma Function**:
The integral \( \int_0^\infty u^{\frac{1}{4}} e^{-u} \, du \) is related to the Gamma function:
\[ \Gamma\left(\frac{5}{4}\right) = \int_0^\infty u^{\frac{1}{4}} e^{-u} \, du \]

7. **Final Expression for N**:
Thus, we have:
\[ N = 4\pi n_0 \cdot \frac{1}{4\alpha^{\frac{3}{4}}} \cdot \Gamma\left(\frac{5}{4}\right) \]
This shows that \( N \) is proportional to \( n_0 \alpha^{-\frac{3}{4}} \).

### Conclusion:
The total number of molecules \( N \) is proportional to:
\[ N \propto n_0 \alpha^{-\frac{3}{4}} \]