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=(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=(h)/(mv)`

Plank's constant is equivalent to:

`h=6.626*10^(-34)Js`

Substituting the values we get:

Solving the operations:

`\lambda=4.73*10^(-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=(\lambda_(electron))/(\lambda)=(2.4*10^(-11))/(4.73*10^(-30))=5.07*10^(18)`

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