In the two vessels of same volume, atomic hydrogen and helium at pressure 1 atm and 2 atm are filled. If temperature of both the sample is same then average speed of hydrogen atoms `ltC_(B)lt` will be related to that of helium `ltC_(He)gt` as

Options

A. `ltC_(H)gt=sqrt(2)ltC_(He)gt`
B. `ltC_(H)gt = ltC_(He)gt`
C. `ltC_(H)gt=2ltC_(He)gt`
D. `ltC_(H)gt=(ltC_(He)gt)/(2)`

Explanation

To solve the problem, we will use the formula for the average speed of gas molecules, which is given by:

\[ C = \sqrt{\frac{8RT}{M}} \]

Where:
- \( C \) is the average speed of the gas molecules,
- \( R \) is the universal gas constant,
- \( T \) is the absolute temperature,
- \( M \) is the molar mass of the gas.

### Step 1: Identify the conditions
We have two gases: atomic hydrogen (H) and helium (He). The conditions provided are:
- Volume of both vessels is the same.
- Pressure of hydrogen is 1 atm and that of helium is 2 atm.
- Temperature \( T \) is the same for both gases.

### Step 2: Write the formula for average speed for both gases
For hydrogen:
\[ C_H = \sqrt{\frac{8RT}{M_H}} \]

For helium:
\[ C_{He} = \sqrt{\frac{8RT}{M_{He}}} \]

### Step 3: Relate the average speeds
Since the temperature \( T \) and the gas constant \( R \) are the same for both gases, we can compare the average speeds directly:

\[ \frac{C_H}{C_{He}} = \sqrt{\frac{M_{He}}{M_H}} \]

### Step 4: Find the molar masses
The molar mass of hydrogen \( M_H \) is approximately 1 g/mol, and the molar mass of helium \( M_{He} \) is approximately 4 g/mol.

### Step 5: Substitute the molar masses into the equation
Substituting the values of molar masses into the equation gives:

\[ \frac{C_H}{C_{He}} = \sqrt{\frac{4}{1}} = \sqrt{4} = 2 \]

### Step 6: Conclusion
Thus, we find that:

\[ C_H = 2 C_{He} \]

This means that the average speed of hydrogen atoms is twice that of helium atoms.

### Final Answer:
The average speed of hydrogen atoms \( C_H \) is related to that of helium \( C_{He} \) as:

\[ C_H = 2 C_{He} \]