Question

You decide to roll a 0.14-kg ball across the floor so slowly that it will have a small momentum and a large de Broglie wavelength. If you roll at 1.0x10^-3 m/s, what is it’s wavelength? How does this compare with the de Broglie wavelength of the high-speed electron that strikes the back face of one of the early models of a TV screen at 1/10 the speed of light (2.4x10^-11 m)?

Answer 1

We are asked to determine the wavelength of a 0.14 kg object with a speed of 1x10^-3 m/s. To do that we will use the following formula:

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

Where:

$\begin{gathered} \lambda=\text{ wavelength} \\ h=\text{ Plank's constant} \\ p=\text{ momentum} \end{gathered}$

The momentum "p" is given by:

$p=mv$

Where:

$\begin{gathered} m=\text{ mass} \\ v=\text{ velocity} \end{gathered}$

Substituting the momentum in the formula for the wavelength:

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

Plank's constant is equivalent to:

$h=6.626\times10^{-34}Js$

Substituting the values we get:

$\lambda=\frac{6.626\times10^{-34}Js}{(0.14\operatorname{kg})(1\times10^{-3}\frac{m}{s})}$

Solving the operations:

$\lambda=4.73\times10^{-30}m$

To compare this wavelength with the wavelength of a high-speed electron moving 1/10 of the speed of light we will determine the quotient between the two wavelengths:

$r=\frac{\lambda_{electron}}{\lambda}=\frac{2.4\times10^{-11}}{4.73\times10^{-30}}=5.07\times10^{18}$

Therefore, the wavelength of the electron is 5.07x10^18 times larger than the wavelength of the ball.