The rms speed of hydrogen molecule at a certain temperature is v. If the temperature is doubled and hydrogen gas dissociates into atomic hydrogen , the rms speed will become

Options

A. v
B. `v/2`
C. `2v`
D. `sqrt2v`

Answer

To solve the problem, we need to determine the new root mean square (rms) speed of hydrogen atoms after the temperature is doubled and the hydrogen molecules dissociate into atomic hydrogen.

### Step-by-Step Solution:

1. **Understanding the rms Speed Formula**:
The rms speed (Vrms) of a gas is given by the formula:
\[ V_{\text{rms}} = \sqrt{\frac{3RT}{M}} \]
where:
- \( R \) is the universal gas constant,
- \( T \) is the absolute temperature,
- \( M \) is the molar mass of the gas.

2. **Initial Conditions**:
- Let the initial rms speed of hydrogen molecules (H₂) at temperature \( T_1 \) be \( V \).
- The molar mass of hydrogen molecule (H₂) is \( 2 \times 10^{-3} \, \text{kg/mol} \).

3. **Doubling the Temperature**:
When the temperature is doubled, the new temperature \( T_2 \) becomes:
\[ T_2 = 2T_1 \]

4. **Dissociation of Hydrogen Molecules**:
When hydrogen gas (H₂) dissociates into atomic hydrogen (H), the molar mass of atomic hydrogen is:
\[ M_{\text{H}} = 1 \times 10^{-3} \, \text{kg/mol} \]

5. **Calculating New rms Speed**:
Now we need to find the new rms speed \( V_2 \) for the atomic hydrogen at the new temperature \( T_2 \):
\[ V_2 = \sqrt{\frac{3R(2T_1)}{M_{\text{H}}}} \]
Substituting the values:
\[ V_2 = \sqrt{\frac{3R(2T_1)}{1 \times 10^{-3}}} \]

6. **Relating to Initial rms Speed**:
We know from the initial conditions that:
\[ V = \sqrt{\frac{3RT_1}{2 \times 10^{-3}}} \]
To find \( V_2 \) in terms of \( V \):
\[ V_2 = \sqrt{2} \cdot \sqrt{\frac{3RT_1}{1 \times 10^{-3}}} \]
Simplifying further, we can express \( V_2 \) as:
\[ V_2 = \sqrt{2} \cdot \sqrt{2} \cdot V = 2V \]

7. **Conclusion**:
Therefore, the new rms speed of atomic hydrogen after the temperature is doubled and the molecules dissociate is:
\[ V_2 = 2V \]

### Final Answer:
The rms speed will become \( 2V \).