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?

1 Answer
Dec 19, 2016

The de Broglie wavelength of an object is given by lambda = h/pλ=hp. With some unit conversions, for this bullet lambda = 1.1 times 10^-33λ=1.1×1033 mm.

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 1000/(60 times 60)100060×60 to convert to ms^-1ms1: 62.59 ms^-1ms1.

The momentum of the bullet, pp, will then be:

p = mv = 0.00957p=mv=0.00957 kg times 62.59kg×62.59 ms^-1 = 0.599ms1=0.599 kgms^-1kgms1

Then lambda = h/pλ=hp where hh is Planck's constant, 6.63 times 10^-346.63×1034 m^2kg^-1.m2kg1.

lambda = (6.63 times 10^-34)/0.599 = 1.1 times 10^-33λ=6.63×10340.599=1.1×1033 mm

This is a very, very tiny wavelength indeed: for comparison, the wavelength of light is only on the order of 10^-7107 mm. For all practical purposes, the wave nature of the bullet is undetectable.