What change in molar energy (in J) would be associated with an atomic transition giving rise to a radiation of 1 Hz frequency: `(h=66xx10^(-34)` Js) Report your answer after multiplying by `10^10`

Answer

To find the change in molar energy associated with an atomic transition giving rise to radiation of 1 Hz frequency, we can use the formula:

\[ \Delta E = h \nu \]

Where:
- \(\Delta E\) is the change in energy,
- \(h\) is Planck's constant (\(6.626 \times 10^{-34} \, \text{Js}\)),
- \(\nu\) is the frequency of the radiation (in this case, \(1 \, \text{Hz}\)).

### Step-by-Step Solution:

1. **Identify the values**:
- Planck's constant, \(h = 6.626 \times 10^{-34} \, \text{Js}\)
- Frequency, \(\nu = 1 \, \text{Hz}\)

2. **Calculate the change in energy for one photon**:
\[ \Delta E = h \nu = (6.626 \times 10^{-34} \, \text{Js}) \times (1 \, \text{Hz}) = 6.626 \times 10^{-34} \, \text{J} \]

3. **Convert the energy from per photon to per mole**:
To convert the energy per photon to energy per mole, we multiply by Avogadro's number (\(N_A = 6.022 \times 10^{23} \, \text{mol}^{-1}\)):
\[ \Delta E_{\text{mole}} = \Delta E \times N_A = (6.626 \times 10^{-34} \, \text{J}) \times (6.022 \times 10^{23} \, \text{mol}^{-1}) \]

4. **Perform the multiplication**:
\[ \Delta E_{\text{mole}} = 6.626 \times 10^{-34} \times 6.022 \times 10^{23} \approx 3.986 \times 10^{-10} \, \text{J/mol} \]

5. **Multiply by \(10^{10}\)**:
To report the answer after multiplying by \(10^{10}\):
\[ \Delta E_{\text{mole}} \times 10^{10} \approx 3.986 \times 10^{-10} \times 10^{10} = 3.986 \, \text{J/mol} \]

### Final Answer:
The change in molar energy associated with the atomic transition is approximately **4 J/mol** (after rounding).