Let the most probable velocity of hydrogen molecules at a temperature `t^(@)C` is `V_(0)`. Suppose all the molecules dissociates into atoms when temperature is raised to `(2t+273).^(@)C` then the new rms velocity is

Options

A. `sqrt(2//3)V_0`
B. `sqrt(3(2 + 273//t) V_0`
C. `2sqrt3V_0`
D. `sqrt6 V_0`

Answer

To find the new RMS velocity of hydrogen atoms after the dissociation of hydrogen molecules at a raised temperature, we can follow these steps:

### Step 1: Understand the given information
We know that the most probable velocity of hydrogen molecules at temperature \( T \) °C is \( V_0 \). When the temperature is raised to \( (2T + 273) \) °C, the hydrogen molecules dissociate into hydrogen atoms.

### Step 2: Convert the temperature to Kelvin
The temperature in Kelvin is given by:
\[ T(K) = T(°C) + 273 \]
So, at the initial temperature \( T \) °C, the temperature in Kelvin is:
\[ T + 273 \]
At the new temperature \( (2T + 273) \) °C, the temperature in Kelvin is:
\[ (2T + 273) + 273 = 2T + 546 \]

### Step 3: Use the formula for most probable velocity
The most probable velocity \( V_{mp} \) of gas molecules is given by:
\[ V_{mp} = \sqrt{\frac{2RT}{m}} \]
Where:
- \( R \) is the universal gas constant,
- \( T \) is the absolute temperature in Kelvin,
- \( m \) is the molar mass of the gas.

For hydrogen molecules (H₂), the molar mass \( m \) is 2 g/mol.

At the initial temperature \( T + 273 \):
\[ V_0 = \sqrt{\frac{2R(T + 273)}{2}} = \sqrt{R(T + 273)} \]

### Step 4: Calculate the new most probable velocity after dissociation
After dissociation, the hydrogen atoms (H) have a molar mass of 1 g/mol. The new temperature is \( 2T + 546 \) K. Therefore, the new most probable velocity \( V_{mp}' \) is:
\[ V_{mp}' = \sqrt{\frac{2R(2T + 546)}{1}} = \sqrt{2R(2T + 546)} \]

### Step 5: Relate the new most probable velocity to \( V_0 \)
Now, we can express the new most probable velocity in terms of \( V_0 \):
\[ V_{mp}' = \sqrt{2R(2T + 546)} = \sqrt{2R(2(T + 273))} = \sqrt{4R(T + 273)} = 2\sqrt{R(T + 273)} = 2V_0 \]

### Step 6: Calculate the new RMS velocity
The RMS velocity \( V_{rms} \) is given by:
\[ V_{rms} = \sqrt{\frac{3RT}{m}} \]
For hydrogen atoms at the new temperature:
\[ V_{rms}' = \sqrt{\frac{3R(2T + 546)}{1}} = \sqrt{3R(2T + 546)} \]

### Step 7: Substitute \( T + 273 \) in terms of \( V_0 \)
Using the relationship we derived earlier:
\[ V_{rms}' = \sqrt{3R(2T + 546)} = \sqrt{3R(2(T + 273))} = \sqrt{6R(T + 273)} = \sqrt{6}V_0 \]

### Final Answer
Thus, the new RMS velocity after the dissociation of hydrogen molecules into atoms at the raised temperature is:
\[ V_{rms}' = \sqrt{6} V_0 \]