# Orbital Angular Momentum Quantum Number (\( l \)) for the Moon We are to find the orbital angular momentum quantum number (\( l \)) for the Moon when it is at a distance of \( r = 3.69 \times 10^8 \) m from Earth. ## Step 1: Classical Angular Momentum The **classical angular momentum** of the Moon in its orbit is: \[ L = m v r \] Where: - \( m \) = mass of Moon (\( 7.35 \times 10^{22 \) kg) - \( v \) = orbital speed of Moon - \( r \) = orbital radius (\(3.69 \times 10^8 \) m) The Moon’s average orbital speed is about \( v = 1.02 \times 103 \) m/s. Plug in values: \[ L = (7.35 \times 10^{22} \text{ kg}) \cdot (1.02 \times 10^3 \text{ m/s}) \cdot (3.69 \times 10^8 \text{ m}) \] \[ L \approx 2.77 \times 10^{34} \text{ kg m}^2/\text{s} \] ## Step 2: Quantization of Angular Momentum In quantum mechanics, the angular momentum is quantized: \[ L = \sqrt{l(l+1)} \hbar \] For very large \( l \), \( \sqrt{l(l+1)} \approx l \). Thus, \[ l \approx \frac{L}{\hbar} \] Where \( \hbar = 1.055 \times 10^{-34} \) J·s. ## Step 3: Calculate \( l \) \[ l \approx \frac{2.77 \times 10^{34}}{1.055 \times 10^{-34}} \] \[ l \approx 2.63 \times 10^{68} \] ## Step 4: Interpretation - The quantum number \( l \) for the Moon is **enormously large**: \( \approx 2.6 \times 10^{68} \). - The difference between adjacent quantum levels (\( \Delta l = 1 \)) is completely unobservable for such macroscopic systems. --- **Summary Table** | Quantity | Value | |------------------|----------------------------| | Distance \( r \) | \( 3.69 \times 10^8 \) m | | Mass \( m \) | \( 7.35 \times 10^{22} \) kg| | Speed \( v \) | \( 1.02 \times 10^3 \) m/s | | \( \hbar \) | \( 1.055 \times 10^{-34} \) J·s | | Quantum \( l \) | \( \approx 2.6 \times 10^{68} \) | --- ## Conclusion **The orbital angular momentum quantum number \( l \) for the Moon is approximately \( 2.6 \times 10^{68} \). For macroscopic objects, the quantization of angular momentum is indistinguishable from a continuum.**