# A certain rifle bullet has a mass of 9.57 g. What is the de Broglie wavelength of the bullet traveling at 140 miles per hour?

Dec 19, 2016

The de Broglie wavelength of an object is given by $\lambda = \frac{h}{p}$. With some unit conversions, for this bullet $\lambda = 1.1 \times {10}^{-} 33$ $m$.

#### Explanation:

Louis de Broglie's key insight was that it is not just light that has a wave and a particle nature, but that other things such as electrons that we normally think of as particles also have a wave nature. For large objects this is usually undetectable, but it is still interesting to calculate, say, the de Broglie wavelength of an elephant.

In this case, we first need to convert the velocity from miles per hour to the SI unit, metres per second. Perhaps first convert to kilometres per hour. One mile is 1.60934 kilometres. We need to multiply 140 mph by this, which yields 225.3 km/h. Then multiply by $\frac{1000}{60 \times 60}$ to convert to $m {s}^{-} 1$: 62.59 $m {s}^{-} 1$.

The momentum of the bullet, $p$, will then be:

$p = m v = 0.00957$ $k g \times 62.59$ $m {s}^{-} 1 = 0.599$ $k g m {s}^{-} 1$

Then $\lambda = \frac{h}{p}$ where $h$ is Planck's constant, $6.63 \times {10}^{-} 34$ ${m}^{2} k {g}^{-} 1.$

$\lambda = \frac{6.63 \times {10}^{-} 34}{0.599} = 1.1 \times {10}^{-} 33$ $m$

This is a very, very tiny wavelength indeed: for comparison, the wavelength of light is only on the order of ${10}^{-} 7$ $m$. For all practical purposes, the wave nature of the bullet is undetectable.