If `H_(2)(g)=2H(g),DeltaH=104cal` , then heat of atomisation of hydrogen is

Options

A. `52Kcal`
B. `104cal`
C. `208Kcal`
D. None of these

Explanation

To solve the problem, we need to understand the concept of heat of atomization and how it relates to the given reaction.

### Step-by-Step Solution:

1. **Identify the Reaction**:
The reaction provided is:
\[ H_2(g) \rightarrow 2H(g) \]
This indicates that one molecule of hydrogen gas (H2) is dissociating into two hydrogen atoms (H).

2. **Understand the Change in Enthalpy (ΔH)**:
The problem states that the change in enthalpy (ΔH) for the reaction is 104 calories. This value represents the energy required to break the bond in the H2 molecule to form two separate hydrogen atoms.

3. **Relate ΔH to Heat of Atomization**:
The heat of atomization is defined as the energy required to convert one mole of a substance into its individual atoms in the gaseous state. In this case, the reaction shows that one mole of H2 produces two moles of H atoms.

4. **Calculate the Heat of Atomization**:
Since the reaction produces 2 moles of hydrogen atoms from 1 mole of H2, the heat of atomization for one mole of hydrogen atoms can be calculated as follows:
\[ \text{Heat of atomization of H} = \frac{\Delta H}{\text{number of moles of H produced}} = \frac{104 \text{ cal}}{2} = 52 \text{ cal} \]

5. **Conclusion**:
Therefore, the heat of atomization of hydrogen is 52 calories.

### Final Answer:
The heat of atomization of hydrogen is **52 calories**.