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 solve the problem of calculating the quantum efficiency for the photochemical reaction where \( A \) converts to \( B \), we can follow these steps:

### Step 1: Understand the given data
- 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

### Step 2: Convert moles of \( B \) to molecules
To find the number of molecules of \( B \) formed, we use Avogadro's number:
\[ \text{Number of molecules of } B = \text{moles of } B \times N_A \]
\[ \text{Number of molecules of } B = 1.0 \times 10^{-5} \text{ moles} \times 6.0 \times 10^{23} \text{ molecules/mole} \]
\[ \text{Number of molecules of } B = 6.0 \times 10^{18} \text{ molecules} \]

### Step 3: Use the formula for quantum efficiency
Quantum efficiency (\( \Phi \)) is defined as the ratio of the number of molecules formed to the number of photons absorbed:
\[ \Phi = \frac{\text{Number of molecules of } B}{\text{Number of photons absorbed}} \]
Substituting the values we have:
\[ \Phi = \frac{6.0 \times 10^{18} \text{ molecules}}{1.2 \times 10^{19} \text{ photons}} \]

### Step 4: Calculate quantum efficiency
Now, we perform the division:
\[ \Phi = \frac{6.0}{1.2} \times 10^{18 - 19} \]
\[ \Phi = 0.5 \times 10^{-1} = 0.5 \]

### Final Answer
The quantum efficiency is \( 0.5 \).

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