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Chapter 7: Problem 42

What is the de Broglie wavelength (in \(\mathrm{nm}\) ) associated with a 2.5 -g Ping-Pong ball traveling 35 mph?

Short Answer

Expert verified
The de Broglie wavelength associated with a 2.5-g Ping-Pong ball traveling at 35 mph is approximately \(1.694 \times 10^{-34}\) nm.

Step by step solution

01

Conversion from mph to m/s

First, convert the speed of the ball from miles per hour (mph) to meters per second (m/s) for consistency with SI units. There are approximately 0.44704 m/s in 1 mph, so multiplying 35 mph by 0.44704 gives approximately \(15.6464 \, m/s \).
02

Calculation of the ping-pong ball's momentum

The momentum (p) of a particle is calculated as the product of its mass and its velocity. The mass must be in kilograms for these calculations to work with standard units. So, convert grams to kilograms by dividing by 1000. So, \(2.5g = 0.0025kg\). Therefore, the formula \(p = m \cdot v\) gives us \(p = 0.0025kg \cdot 15.6464m/s = 0.039116kg \cdot m/s \).
03

Calculation of the de Broglie wavelength

Finally, calculate the de Broglie wavelength using the formula \(\lambda = h / p\) where \(h\) is the Planck constant, equal to \(6.62607015 \times 10^{-34} m^{2}kg/s\). Substituting the computed value of momentum, get \(\lambda = \frac{6.62607015 \times 10^{-34} m^{2}kg/s}{0.039116kg \cdot m/s}\). This value will be in meters, which we convert to nanometers (nm) by multiplying by \(10^{9}\).

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