For a monatomic gas, kinetic energy `= E`. The relation with `rms` velocity is

Options

A. `u = ((3E)/(2m))^(1//2)`
B. `u = ((E)/(2m))^(1//2)`
C. `u = ((E)/(3m))^(1//2)`
D. `u = ((2E)/(m))^(1//2)`

Answer

To find the relationship between the kinetic energy \( E \) of a monatomic gas and its root mean square (rms) velocity \( v_{rms} \), we can follow these steps:

### Step 1: Understand the formulas involved
The kinetic energy \( E \) for a monatomic gas is given by:
\[ E = \frac{3}{2} RT \]
where \( R \) is the gas constant and \( T \) is the temperature in Kelvin.

The root mean square velocity \( v_{rms} \) is given by:
\[ v_{rms} = \sqrt{\frac{3RT}{M}} \]
where \( M \) is the molar mass of the gas.

### Step 2: Relate kinetic energy to temperature
From the kinetic energy formula, we can express \( RT \) in terms of \( E \):
\[ RT = \frac{2}{3} E \]

### Step 3: Substitute \( RT \) in the rms velocity formula
Now, we can substitute \( RT \) into the rms velocity formula:
\[ v_{rms} = \sqrt{\frac{3RT}{M}} = \sqrt{\frac{3 \cdot \frac{2}{3} E}{M}} \]

### Step 4: Simplify the expression
This simplifies to:
\[ v_{rms} = \sqrt{\frac{2E}{M}} \]

### Step 5: Final expression
Thus, we find the relationship between the kinetic energy \( E \) and the rms velocity \( v_{rms} \):
\[ v_{rms} = \sqrt{\frac{2E}{M}} \]

### Conclusion
The final expression shows that the root mean square velocity is proportional to the square root of the kinetic energy divided by the molar mass.