For a photochemical reaction `A to B, 1.0xx10^(-5)` mole of B were formed on absorbing `1.2xx10^(19)` photons, quantum efficiency is : `(N_(A)=6xx10^(23))`

Explanation

To calculate the quantum efficiency for the photochemical reaction \( A \to B \), we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Concept of Quantum Efficiency**:
Quantum efficiency (\( \Phi \)) is defined as the ratio of the number of molecules formed to the number of photons absorbed. The formula is given by:
\[ \Phi = \frac{\text{Number of molecules formed}}{\text{Number of photons absorbed}} \]

2. **Identify the Given Values**:
- Moles of \( B \) formed: \( 1.0 \times 10^{-5} \) moles
- Number of photons absorbed: \( 1.2 \times 10^{19} \) photons
- Avogadro's number (\( N_A \)): \( 6.0 \times 10^{23} \) molecules/mole

3. **Convert Moles of \( B \) to Molecules**:
To find the number of molecules of \( B \) formed, we multiply the number of moles by Avogadro's number:
\[ \text{Number of molecules of } B = 1.0 \times 10^{-5} \text{ moles} \times 6.0 \times 10^{23} \text{ molecules/mole} \]
\[ = 6.0 \times 10^{18} \text{ molecules} \]

4. **Substitute Values into the Quantum Efficiency Formula**:
Now we can substitute the values into the quantum efficiency formula:
\[ \Phi = \frac{6.0 \times 10^{18} \text{ molecules}}{1.2 \times 10^{19} \text{ photons}} \]

5. **Calculate Quantum Efficiency**:
Performing the division:
\[ \Phi = \frac{6.0}{1.2} \times \frac{10^{18}}{10^{19}} = 0.5 \]

6. **Final Result**:
Therefore, the quantum efficiency of the reaction is:
\[ \Phi = 0.5 \]