Two masses are connected with spring constant $k$. The masses have magnitudes $m$ and $M$. The center-of-mass of the system is fixed. If $ k = \text {100 N/m} $ and $m=\dfrac{1}{2}M=\text{1 kg}$, let the ground state energy of the system be $E$. If $E$ can be expressed in the form $ a \times 10^p $ eV (electron-volts), find the ordered pair $(a,p)$, where $ 0 < a < 10 $, and it is rounded to the nearest positive integer and $p$ is an integer. For example, $ 4.2 \times 10^7 $ should be expressed as $(4,7)$. 

[i](Trung Phan, 10 points)[/i]
1. **Determine the reduced mass $\mu$:**
   Given \( m = 1 \, \text{kg} \) and \( M = 2m = 2 \, \text{kg} \), the reduced mass $\mu$ is calculated as:
   \[
   \mu = \frac{mM}{m + M} = \frac{1 \times 2}{1 + 2} = \frac{2}{3} \, \text{kg}
   \]

2. **Express the total energy $E$:**
   The total energy of the system can be written as:
   \[
   E = \frac{1}{2} kx^2 + \frac{1}{2} \mu v^2
   \]
   Using the Heisenberg Uncertainty Principle, we approximate:
   \[
   \Delta x \cdot \Delta p \ge \frac{\hbar}{2}
   \]
   Assuming \(\Delta x \sim x\) and \(\Delta p \sim p\), we get:
   \[
   xp \sim \frac{\hbar}{2}
   \]
   Therefore:
   \[
   p = \frac{\hbar}{2x}
   \]

3. **Substitute $p$ into the energy expression:**
   \[
   E = \frac{1}{2} \left( kx^2 + \frac{p^2}{\mu} \right) = \frac{1}{2} \left( kx^2 + \frac{\left( \frac{\hbar}{2x} \right)^2}{\mu} \right) = \frac{1}{2} \left( kx^2 + \frac{\hbar^2}{4\mu x^2} \right)
   \]

4. **Minimize the energy using the AM-GM inequality:**
   \[
   E = \frac{1}{2} \left( kx^2 + \frac{\hbar^2}{4\mu x^2} \right) \ge \sqrt{kx^2 \cdot \frac{\hbar^2}{4\mu x^2}} = \sqrt{\frac{k\hbar^2}{4\mu}}
   \]
   Therefore, the minimum energy is:
   \[
   E_{\text{min}} = \frac{\hbar}{2} \sqrt{\frac{k}{\mu}}
   \]

5. **Substitute the given values:**
   \[
   k = 100 \, \text{N/m}, \quad \mu = \frac{2}{3} \, \text{kg}, \quad \hbar \approx 1.0545718 \times 10^{-34} \, \text{Js}
   \]
   \[
   E_{\text{min}} = \frac{1.0545718 \times 10^{-34}}{2} \sqrt{\frac{100}{\frac{2}{3}}} = \frac{1.0545718 \times 10^{-34}}{2} \sqrt{150} = \frac{1.0545718 \times 10^{-34}}{2} \times 12.247
   \]
   \[
   E_{\text{min}} \approx 6.45 \times 10^{-34} \, \text{J}
   \]

6. **Convert the energy to electron-volts:**
   \[
   1 \, \text{eV} = 1.602176634 \times 10^{-19} \, \text{J}
   \]
   \[
   E_{\text{min}} \approx \frac{6.45 \times 10^{-34}}{1.602176634 \times 10^{-19}} \, \text{eV} \approx 4.03 \times 10^{-15} \, \text{eV}
   \]

7. **Express the energy in the form \(a \times 10^p\):**
   \[
   E_{\text{min}} \approx 4.03 \times 10^{-15} \, \text{eV}
   \]
   Rounding \(a\) to the nearest positive integer, we get \(a = 4\) and \(p = -15\).

The final answer is \(\boxed{(4, -15)}\).