Calculate energy of 2mole of photons of radiation whose frequency is 5x10^14 Hz

Explanation

To calculate the energy of 2 moles of photons of radiation whose frequency is \(5 \times 10^{14}\) Hz, we can follow these steps:

### Step 1: Calculate the energy of one photon
The energy of a single photon can be calculated using the formula:
\[ E = h \nu \]
where:
- \(E\) is the energy of one photon,
- \(h\) is Planck's constant (\(6.626 \times 10^{-34} \, \text{J s}\)),
- \(\nu\) is the frequency of the radiation (\(5 \times 10^{14} \, \text{s}^{-1}\)).

### Step 2: Substitute the values into the formula
Substituting the values into the formula:
\[ E = (6.626 \times 10^{-34} \, \text{J s}) \times (5 \times 10^{14} \, \text{s}^{-1}) \]

### Step 3: Perform the multiplication
Calculating the above expression:
\[ E = 6.626 \times 5 \times 10^{-34} \times 10^{14} \]
\[ E = 33.13 \times 10^{-20} \, \text{J} \]
\[ E = 3.313 \times 10^{-19} \, \text{J} \]

### Step 4: Calculate the energy for 2 moles of photons
To find the energy for 2 moles of photons, we use Avogadro's number (\(N_A = 6.022 \times 10^{23} \, \text{mol}^{-1}\)):
\[ \text{Energy for 2 moles} = E \times 2 \times N_A \]
Substituting the values:
\[ \text{Energy for 2 moles} = (3.313 \times 10^{-19} \, \text{J}) \times (2 \times 6.022 \times 10^{23} \, \text{mol}^{-1}) \]

### Step 5: Calculate the total energy
Calculating the above expression:
\[ \text{Energy for 2 moles} = 3.313 \times 10^{-19} \times 12.044 \times 10^{23} \]
\[ \text{Energy for 2 moles} = 3.99 \times 10^{4} \, \text{J} \]
\[ \text{Energy for 2 moles} = 3990.2 \, \text{kJ} \]

### Final Answer
The energy of 2 moles of photons of radiation whose frequency is \(5 \times 10^{14}\) Hz is approximately \(3990.2 \, \text{kJ}\).

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