A hydrogen molecule in excited state travelling in the x-direction with kinetic energy of 1.0 eV dissociate into two hydrogen atoms. If one of themn moves perpendicular to the x-direction with a kinetic energy 1 eV. Then the energy released in te dissociation reaction will be

Options

A. 2.6 eV
B. 2 eV
C. 3.6 eV
D. 3 eV

Explanation

To solve the problem, we need to determine the energy released during the dissociation of a hydrogen molecule into two hydrogen atoms, given their kinetic energies.

### Step-by-Step Solution:

1. **Understanding the Initial Conditions:**
- A hydrogen molecule (H₂) in an excited state has a kinetic energy (KE_initial) of 1.0 eV.
- The molecule dissociates into two hydrogen atoms (H) after traveling in the x-direction.

2. **Kinetic Energy of the Atoms After Dissociation:**
- One of the hydrogen atoms moves perpendicular to the x-direction with a kinetic energy (KE_1) of 1.0 eV.
- Let’s denote the second hydrogen atom's kinetic energy as KE_2.

3. **Conservation of Momentum:**
- The initial momentum of the hydrogen molecule (P_initial) can be calculated using its kinetic energy:
\[ P_{\text{initial}} = \sqrt{2 \cdot m \cdot KE_{\text{initial}}} = \sqrt{2 \cdot 2m \cdot 1 \text{ eV}} = \sqrt{4m} = 2\sqrt{m} \]
- The momentum of the first hydrogen atom (moving in the y-direction) is:
\[ P_1 = \sqrt{2 \cdot m \cdot KE_1} = \sqrt{2 \cdot m \cdot 1 \text{ eV}} = \sqrt{2m} \]

4. **Finding the Momentum of the Second Atom:**
- By conservation of momentum:
\[ P_{\text{initial}} = P_1 + P_2 \]
- Rearranging gives:
\[ P_2 = P_{\text{initial}} - P_1 = 2\sqrt{m} - \sqrt{2m} \]

5. **Calculating the Magnitude of P2:**
- The magnitude of \( P_2 \) can be calculated using the Pythagorean theorem since the two momenta are perpendicular:
\[ |P_2| = \sqrt{(2\sqrt{m})^2 + (\sqrt{2m})^2} = \sqrt{4m + 2m} = \sqrt{6m} \]

6. **Finding the Kinetic Energy of the Second Atom:**
- The kinetic energy of the second hydrogen atom (KE_2) can be calculated as:
\[ KE_2 = \frac{|P_2|^2}{2m} = \frac{(6m)}{2m} = 3 \text{ eV} \]

7. **Calculating the Total Final Kinetic Energy:**
- The total final kinetic energy (KE_final) of the two hydrogen atoms is:
\[ KE_{\text{final}} = KE_1 + KE_2 = 1 \text{ eV} + 3 \text{ eV} = 4 \text{ eV} \]

8. **Calculating the Energy Released in the Dissociation:**
- The energy released during the dissociation (ΔE) is given by:
\[ \Delta E = KE_{\text{final}} - KE_{\text{initial}} = 4 \text{ eV} - 1 \text{ eV} = 3 \text{ eV} \]

### Final Answer:
The energy released in the dissociation reaction is **3 eV**.