Energy of 1 mole of radio wave photons with a frequency of
909 KHz is :-

Options

A. `6.02xx10^(-28)J`
B. `3.62xx10^(-4)J`
C. `1.00xx10^(-4)J`
D. `6.02xx10^(-3)J`

Explanation

To find the energy of 1 mole of radio wave photons with a frequency of 909 KHz, we will follow these steps:

### Step-by-Step Solution:

1. **Convert Frequency to Hertz:**
The frequency given is in kilohertz (KHz). To convert it to hertz (Hz), we multiply by \(10^3\).
\[ \text{Frequency} (\nu) = 909 \, \text{KHz} = 909 \times 10^3 \, \text{Hz} = 909000 \, \text{Hz} \]

2. **Identify Constants:**
We need to use the Planck constant (H) and Avogadro's number (N).
- Planck's constant \(H = 6.626 \times 10^{-34} \, \text{Joule second}\)
- Avogadro's number \(N = 6.022 \times 10^{23} \, \text{mol}^{-1}\)

3. **Calculate Energy for One Photon:**
The energy (E) of a single photon can be calculated using the formula:
\[ E = H \nu \]
Substituting the values:
\[ E = 6.626 \times 10^{-34} \, \text{Joule second} \times 909000 \, \text{Hz} \]
\[ E \approx 6.626 \times 10^{-34} \times 909000 \approx 6.022 \times 10^{-27} \, \text{Joules} \]

4. **Calculate Energy for One Mole of Photons:**
To find the energy for one mole of photons, we multiply the energy of one photon by Avogadro's number:
\[ E_{\text{mole}} = N \times E \]
\[ E_{\text{mole}} = 6.022 \times 10^{23} \, \text{mol}^{-1} \times 6.022 \times 10^{-27} \, \text{Joules} \]
\[ E_{\text{mole}} \approx 3.62 \times 10^{-4} \, \text{Joules} \]

5. **Final Answer:**
The energy of 1 mole of radio wave photons with a frequency of 909 KHz is approximately:
\[ \boxed{3.62 \times 10^{-4} \, \text{Joules}} \]