Calculating de-Broglie Wavelength for a Ball

This problem involves calculating the de-Broglie wavelength ($\lambda$) of a macroscopic object (a ball). The de-Broglie hypothesis states that matter particles exhibit wave-like properties. The wavelength associated with a particle is inversely proportional to its momentum.

De-Broglie Wavelength Formula

The formula used to calculate the de-Broglie wavelength is:

$ \lambda = \frac{h}{p} $

where:

• $ \lambda $ is the de-Broglie wavelength (in meters, m)
• $ h $ is Planck's constant, which is approximately $ 6.626 \times 10^{-34} $ J·s (Joule-seconds)
• $ p $ is the momentum of the particle (in kg·m/s)

Momentum ($ p $) is calculated as the product of mass ($ m $) and velocity ($ v $):

$ p = mv $

So, the de-Broglie wavelength formula can also be written as:

$ \lambda = \frac{h}{mv} $

Given Information

• Mass of the ball, $ m = 150 $ g
• Velocity of the ball, $ v = 30.0 $ m/s
• Planck's constant, $ h = 6.626 \times 10^{-34} $ J·s

Step-by-Step Calculation

1. Convert Mass to Kilograms:

   The mass needs to be in the standard SI unit, kilograms (kg).

   $ m = 150 \text{ g} = \frac{150}{1000} \text{ kg} = 0.150 \text{ kg} $

2. Calculate Momentum:

   Use the formula $ p = mv $.

   $ p = (0.150 \text{ kg}) \times (30.0 \text{ m/s}) $

   $ p = 4.5 \text{ kg·m/s} $

3. Calculate de-Broglie Wavelength:

   Substitute the values of $ h $ and $ p $ into the formula $ \lambda = \frac{h}{p} $.

   $ \lambda = \frac{6.626 \times 10^{-34} \text{ J·s}}{4.5 \text{ kg·m/s}} $

   Since 1 J = 1 kg·m$^2$/s$^2$, the units become:

   $ \lambda = \frac{6.626 \times 10^{-34} \text{ kg·m}^2/\text{s}}{4.5 \text{ kg·m/s}} $

   $ \lambda \approx 1.4724 \times 10^{-34} \text{ m} $

4. Round to Appropriate Significant Figures:

   The given velocity (30.0 m/s) has three significant figures. The mass (150 g) can be interpreted as having two or three significant figures depending on context, but typically in physics problems like this, trailing zeros after the decimal imply significance (0.150 kg has three). Planck's constant is known to higher precision. Rounding the result to three significant figures:

   $ \lambda \approx 1.47 \times 10^{-34} \text{ m} $

Conclusion

The calculated de-Broglie wavelength associated with the ball is approximately $ 1.47 \times 10^{-34} $ meters. This extremely small wavelength highlights why wave-like properties are not observable for macroscopic objects in everyday life.