The `C_("rms")` of He `5.3 xx 10^(2) ms^(-1)` and the distance per collision is `6.18 xx 10^(-8)` metre `"collision"^(-1)`. The collision frequency is `x xx 10^(9)` collision `sec^(-1)`. What will be the value of x?

Answer

To find the value of \( x \) in the collision frequency equation, we can follow these steps:

### Step 1: Understand the formula for collision frequency
The collision frequency (\( f \)) can be calculated using the formula:
\[ f = \frac{v_{\text{rms}}}{d} \]
where:
- \( v_{\text{rms}} \) is the root mean square speed (given as \( 5.3 \times 10^2 \, \text{m/s} \))
- \( d \) is the distance per collision (given as \( 6.18 \times 10^{-8} \, \text{m} \))

### Step 2: Substitute the values into the formula
Substituting the given values into the formula:
\[ f = \frac{5.3 \times 10^2 \, \text{m/s}}{6.18 \times 10^{-8} \, \text{m}} \]

### Step 3: Perform the division
Now, we will perform the division:
\[ f = \frac{5.3}{6.18} \times \frac{10^2}{10^{-8}} = \frac{5.3}{6.18} \times 10^{2 + 8} = \frac{5.3}{6.18} \times 10^{10} \]

### Step 4: Calculate \( \frac{5.3}{6.18} \)
Calculating \( \frac{5.3}{6.18} \):
\[ \frac{5.3}{6.18} \approx 0.856 \]

### Step 5: Combine the results
Now, substituting back into the frequency equation:
\[ f \approx 0.856 \times 10^{10} \, \text{collisions/sec} \]

### Step 6: Express in terms of \( x \)
Since the collision frequency is given as \( x \times 10^9 \):
\[ 0.856 \times 10^{10} = x \times 10^9 \]

### Step 7: Solve for \( x \)
To find \( x \):
\[ x = 0.856 \times 10^{10} / 10^{9} = 0.856 \times 10 = 8.56 \]

Thus, the value of \( x \) is approximately:
\[ \boxed{8.56} \]