Question #663b0

In order to be able to calculate the de Broglie wavelength of the electron at that particular speed, you must know its mass

`m_"e" ~~ 9.1094 * 10^(-31)"kg"`

Now, the de Broglie wavelength depends on the momentum of the electron, as given by the equation

`color(blue)(ul(color(black)(lamda = h/p))) ->` the de Broglie wavelength

Here

• `p` is the momentum of the electron
• `lamda` is its wavelength
• `h` - Planck's constant, equal to `6.626 * 10^(-34)"kg m"^2"s"^(-1)`

The momentum of the electron can be calculated by using the equation

`color(blue)(ul(color(black)(p = m * v)))`

Here

• `m` is the mass of the particle
• `v` is its velocity

In your case, you have

`p = 9.1094 * 10^(-31)"kg" * "1500 m s"^(-1)`

`p = 1.3664 * 10^(-27)"kg m s"^(-1)`

This means that the de Broglie wavelength associated with the electron at this particular speed is equal to

`lamda = (6.626 * 10^(-34) color(red)(cancel(color(black)("kg"))) "m"^color(red)(cancel(color(black)(2)))color(red)(cancel(color(black)("s"^(-1)))))/(1.3664 * 10^(-27)color(red)(cancel(color(black)("kg")))color(red)(cancel(color(black)("m")))color(red)(cancel(color(black)("s"^(-1)))))`

`color(darkgreen)(ul(color(black)(lamda = 4.8 * 10^(-7)"m" = "480 nm")))`

The answer is rounded to two sig figs.