If the energy difference between the electronic states is `214.68 kJmol^(-1)` calculate the frequency of light emitted when an electron drop form the higher to the lower state planks constant , `h = 39.79 xx 10^(-14)kJmol^(-1)`

Explanation

To solve the problem, we need to calculate the frequency of light emitted when an electron drops from a higher electronic state to a lower one, given the energy difference between the states and Planck's constant.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Energy difference (ΔE) = 214.68 kJ/mol
- Planck's constant (h) = 39.79 x 10^(-14) kJ/mol·s

2. **Use the Formula for Energy of a Photon:**
The energy of a photon can be expressed using Planck's equation:
\[ E = h \nu \]
where:
- \(E\) is the energy of the photon,
- \(h\) is Planck's constant,
- \(\nu\) is the frequency of the photon.

3. **Rearrange the Formula to Solve for Frequency:**
To find the frequency (\(\nu\)), we can rearrange the equation:
\[ \nu = \frac{E}{h} \]

4. **Convert Energy from kJ/mol to kJ/s:**
Since we want the frequency in terms of seconds, we need to convert the energy from kJ/mol to kJ/s. We can do this by dividing the energy by Avogadro's number (\(N_A\)), which is approximately \(6.022 \times 10^{23}\) mol⁻¹:
\[ E = \frac{214.68 \text{ kJ/mol}}{6.022 \times 10^{23} \text{ mol}^{-1}} \approx 3.57 \times 10^{-19} \text{ kJ} \]

5. **Calculate the Frequency:**
Now we can substitute the values into the frequency formula:
\[ \nu = \frac{3.57 \times 10^{-19} \text{ kJ}}{39.79 \times 10^{-14} \text{ kJ/s}} \approx 5.4 \times 10^{14} \text{ s}^{-1} \]

### Final Answer:
The frequency of light emitted when the electron drops from the higher to the lower state is approximately:
\[ \nu \approx 5.4 \times 10^{14} \text{ Hz} \]